Quadratic Graphs: Vertex, Roots and Equation

Who this is for

Use this lesson when your course is studying quadratic relations and their graphs. You should already be able to expand brackets, substitute into a formula and solve a simple linear equation. The graph-and-form work supports selected quadratic-relations expectations from Ontario Grade 10 Academic MPM2D. Completing the square with other leading coefficients extends that work toward selected Grade 11 MCR3U and MCF3M expectations. In Quebec, the quadratic forms and methods required depend on the secondary year and mathematics option. Check your course outline; this is a topic lesson, not a complete provincial course or exam guide.

Already comfortable with roots and graphs? Continue to the optional quadratic inequalities extension and its practice questions.

Working with two graphs? Try the optional arch and straight-segment system, then three new segments.

1. One curve, three useful forms

A quadratic function has the form $y = ax^2 + bx + c$. Here $x$ is the input, $y$ is the output, and $a$, $b$ and $c$ are fixed real coefficients, with $a \ne 0$. Its graph is a parabola. For this lesson, $x$ can be any real number unless a problem gives a narrower domain.

Consider these three expressions:

$$y = x^2 - 6x + 5$$

$$y = (x - 1)(x - 5)$$

$$y = (x - 3)^2 - 4$$

They give the same $y$ for every $x$. Expanding the product gives $x^2 - 5x - x + 5 = x^2 - 6x + 5$. Expanding the square gives $x^2 - 6x + 9 - 4 = x^2 - 6x + 5$.

To find the factored form of this example, look for two numbers whose sum is $-6$ and whose product is 5: $-1$ and $-5$. This works because $(x + p)(x + q)$ expands to $x^2 + (p + q)x + pq$, where $p$ and $q$ are those two numbers. Always expand your factors to check. This shortcut applies directly when the coefficient of $x^2$ is 1.

Each form makes a different feature easier to read:

Expanded form shows the $y$-intercept: when $x = 0$, $y = 5$, so the point is $(0, 5)$.

Factored form shows the roots: $y = 0$ at $x = 1$ and $x = 5$, so the $x$-intercepts are $(1, 0)$ and $(5, 0)$.

Vertex form shows the turning point: $(3, -4)$. The smallest possible value of $(x - 3)^2$ is zero, so the minimum $y$-value is $-4$.

The axis of symmetry is $x = 3$. Inputs equally far from 3 have equal outputs. For example, $x = 2$ and $x = 4$ both give $y = -3$.

Check: Do the roots tell you where the curve crosses the $y$-axis?

2. Build a sketch that agrees with the equation

For $y = x^2 - 6x + 5$, plot $(0, 5)$, $(1, 0)$, $(2, -3)$, $(3, -4)$, $(4, -3)$, $(5, 0)$ and $(6, 5)$. Draw a smooth curve through them. It opens upward because the coefficient of $x^2$ is positive. Label the axes and the vertex. Its domain (the allowed inputs) is all real numbers and its range (the possible outputs) is $y \ge -4$.

Do not connect the points with straight segments and call that the parabola. The points guide the curve. A graphing tool can check your sketch after you have predicted its key features.

The general vertex form $y = a(x - h)^2 + k$ has fixed real parameters $a$, $h$ and $k$, with $a \ne 0$. Here $h$ is the vertex’s $x$-coordinate and $k$ is its $y$-coordinate, so the vertex is $(h, k)$. A positive $a$ gives a minimum; a negative $a$ gives a maximum. The sign inside the brackets needs care: $(x + 2)^2$ means $(x - (-2))^2$, so the vertex's $x$-coordinate is $-2$.

3. Solve the question that was actually asked

For $y = (x - 3)^2 - 4$:

To find the roots, set $y = 0$: $(x - 3)^2 = 4$, so $x - 3 = 2$ or $-2$. Therefore $x = 5$ or 1.

To find where $y = 5$, solve $(x - 3)^2 - 4 = 5$. Then $(x - 3)^2 = 9$, so $x = 6$ or 0.

If a product is zero, at least one factor must be zero. Thus $(x - 1)(x - 5) = 0$ means $x - 1 = 0$ or $x - 5 = 0$; the factors need not both be zero. This zero-product rule applies only when the product is zero. If $(x - 1)(x - 5) = 5$, you cannot set either factor equal to 5 and expect all solutions.

Some parabolas have no real roots. For example, $y = (x - 3)^2 + 2$ is always at least 2, so its graph never reaches the $x$-axis. It still has a vertex and a $y$-intercept.

4. Complete the square to find the vertex and solve equations

Rewrite $x^2 - 6x + 5$ by completing a square:

$$\begin{aligned}x^2 - 6x + 5 &= x^2 - 6x + 9 - 9 + 5 \\&= (x - 3)^2 - 4\end{aligned}$$

Why add 9? Half of $-6$ is $-3$, and $(-3)^2 = 9$. Why subtract 9 too? To preserve the original value. This worked example has leading coefficient 1; when it is different, account for that coefficient before completing the square.

Try: Write $x^2 + 4x + 1$ in vertex form.

Complete the square when the coefficient is different from 1

The same idea works when the coefficient of $x^2$ is different from 1. First factor that coefficient out of the terms containing $x$. Leave the constant outside the brackets for now. Then complete the square inside the brackets.

The detail to watch is the multiplier outside: adding a number inside the brackets changes the whole expression by that number times the multiplier.

Example with a positive coefficient

Write $y=2x^2-12x+10$ in vertex form.

Factor 2 out of the first two terms:

$$y=2(x^2-6x)+10.$$

Inside the brackets, half of $-6$ is $-3$, and $(-3)^2=9$. Add and subtract 9 inside the same brackets:

$$\begin{aligned} y&=2[x^2-6x+9-9]+10\\ &=2[(x-3)^2-9]+10\\ &=2(x-3)^2-18+10\\ &=2(x-3)^2-8. \end{aligned}$$

Why does $-9$ become $-18$? The 2 multiplies every term inside the brackets, including $-9$.

You can also keep the compensation outside from the start:

$$y=2(x^2-6x+9)+10-18.$$

Adding 9 inside adds $2\times9=18$ to the whole expression, so subtract 18 outside. Both approaches preserve the original expression.

The vertex is $(3,-8)$. Since $2(x-3)^2\geq0$, the minimum value of $y$ is $-8$, reached when $x=3$.

Check by expanding:

$$\begin{aligned}&2(x-3)^2-8\\&=2x^2-12x+18-8\\&=2x^2-12x+10.\end{aligned}$$

Example with a negative coefficient

Write $y=-3x^2-12x-9$ in vertex form.

Factor $-3$ out of the first two terms. Notice that $(-3)(4x)=-12x$, so the term inside is $+4x$:

$$y=-3(x^2+4x)-9.$$

Half of 4 is 2, and $2^2=4$. Add and subtract 4 inside:

$$\begin{aligned} y&=-3[x^2+4x+4-4]-9\\ &=-3[(x+2)^2-4]-9\\ &=-3(x+2)^2+12-9\\ &=-3(x+2)^2+3. \end{aligned}$$

Here, the compensation is positive outside because $(-3)(-4)=+12$.

The other way to write the same step is

$$y=-3(x^2+4x+4)-9+12.$$

Adding 4 inside changes the whole expression by $(-3)\times4=-12$. Add 12 outside to balance that change.

The vertex is $(-2,3)$. Since $-3(x+2)^2\leq0$, the maximum value is 3, reached when $x=-2$.

Check by expanding:

$$\begin{aligned}&-3(x+2)^2+3\\&=-3x^2-12x-12+3\\&=-3x^2-12x-9.\end{aligned}$$

A quick check for the compensation

If the brackets have a multiplier $a$, adding $q$ inside changes the whole expression by $aq$. Subtract $aq$ outside to preserve its value. When $a$ is negative, subtracting $aq$ may mean adding a positive number.

You can instead add and subtract $q$ inside the same brackets. When you later distribute $a$, multiply both terms by $a$.

See where the quadratic formula comes from

You can use the formula below without reproducing its derivation. Follow these optional steps to see why it works; check your course outline to see whether you need to reproduce them.

Now use letters in place of the numbers. In

$$ax^2+bx+c=0,$$

$x$ is the unknown, and $a$, $b$ and $c$ are fixed real numbers. The number $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$, and $c$ is the constant term. Their signs are part of their values. We require $a\ne0$; otherwise the equation is not quadratic.

First complete the square in the expression. Here the compensation is written outside the brackets:

$$\begin{aligned} &ax^2+bx+c\\ &=a\left[x^2+\frac ba x\right]+c\\ &=a\left[x^2+\frac ba x+\left(\frac b{2a}\right)^2\right]\\ &\quad-a\left(\frac b{2a}\right)^2+c\\ &=a\left(x+\frac b{2a}\right)^2\\ &\quad-\frac{b^2}{4a}+c\\ &=a\left(x+\frac b{2a}\right)^2\\ &\quad-\frac{b^2-4ac}{4a}. \end{aligned}$$

Inside the brackets, half of $b/a$ is $b/(2a)$. The compensating term outside is $-a[b/(2a)]^2=-b^2/(4a)$, using exactly the same multiplier rule as the examples.

Define $\Delta=b^2-4ac$. This number is called the discriminant. The completed-square form is

$$y=a\left(x+\frac b{2a}\right)^2-\frac{\Delta}{4a}.$$

This also gives the vertex: $\left(-\frac b{2a},-\frac{\Delta}{4a}\right)$.

To find the roots, set $y=0$ and isolate the square:

$$\begin{aligned} a\left(x+\frac b{2a}\right)^2&=\frac{\Delta}{4a}\\ 4a^2\left(x+\frac b{2a}\right)^2&=\Delta\\ (2ax+b)^2&=\Delta. \end{aligned}$$

The second line multiplies both sides by $4a$. This is valid for positive or negative $a$, because $a\ne0$.

If $\Delta\geq0$, a number whose square is $\Delta$ can equal $+\sqrt{\Delta}$ or $-\sqrt{\Delta}$, so

$$2ax+b=\pm\sqrt{\Delta}.$$

Subtract $b$, then divide by $2a$:

$$\boxed{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}.$$

That is the quadratic formula. The symbol $\pm$ tells you to consider both the plus and the minus choice. The denominator $2a$ divides the whole numerator. When the square root is zero, the two choices give the same solution.

Tell how many real solutions there are

Look at $(2ax+b)^2=\Delta$. A square of a real number cannot be negative.

• If $\Delta>0$, there are two distinct real solutions. The graph of $y=ax^2+bx+c$ crosses the $x$-axis at two points.

• If $\Delta=0$, there is one distinct real solution, $x=-b/(2a)$. It is called a double root. The graph touches the $x$-axis at its vertex.

• If $\Delta<0$, there are no real solutions and no $x$-intercepts. This does not mean the quadratic has no graph; its graph simply does not meet the $x$-axis.

These statements apply whether the parabola opens upward or downward. This part of the lesson works with real-number solutions.

Before using the discriminant or the formula, rearrange your equation so one side is zero and identify $a$, $b$ and $c$, including their signs.

Practise in three steps

Use the hints or open a worked answer whenever it helps.

5. Use a quadratic model and respect its domain

A rectangular garden has 20 m of fencing around all four sides. Let its width be $w$ metres and its length be $l$ metres. Ignoring fence thickness, $2w + 2l = 20$, so $l = 10 - w$.

Let $A$ be the numerical value of the area in square metres. Then $$\begin{aligned}A &= w(10 - w) \\&= -w^2 + 10w \\&= -(w - 5)^2 + 25\end{aligned}$$.

A real garden needs $w > 0$ and $l > 0$, so $0 < w < 10$. The graph opens downward and its maximum occurs at $w = 5$. The length is also 5 m, and the maximum area is $25\text{ m}^2$. Check: the perimeter is $2(5) + 2(5) = 20$ m.

The algebraic roots $w = 0$ and $w = 10$ describe zero-area boundary cases. They are not admissible widths for this garden. The model's physical domain matters even though the polynomial can be evaluated outside it. No differentiation is needed: the squared term establishes the maximum.

When a line meets a parabola: solve the system, then check the segment

This optional 10–15 minute route connects linear systems by substitution to quadratic graphs. You will build two rules from a diagram, find their shared points and decide which points belong to the stated objects. Review vertex form and quadratic equations above if needed. If this is not your current topic, continue to the independent quadratic practice.

Use the method your course asks for. Quebec’s progression places first-/second-degree systems in Secondary IV Science (SN), with reinvestment in V, and graphical solutions in Secondary V TS. Ontario MCR3U includes graphical and algebraic linear–quadratic intersections. The exact algebra below is an optional explanation when your course asks for graphical estimates; it is not a requirement for every secondary option. See the curriculum references.

1. Turn the geometric information into two rules

An arch is modelled by a parabola with endpoints $(0,0)$ and $(8,0)$, with highest point $(4,8)$. A straight segment joins $(0,2)$ to $(3,5)$. Where do their centreline models meet? Ignore thickness and structural forces: this is a geometric model.

Let $x$ be horizontal position from the arch’s left endpoint and $y$ height above the ground, both expressed as numerical values in metres. Write $A(x)$ for the arch’s height and $B(x)$ for the segment’s height at position $x$. Here $A$ names a height function, not the garden’s area from the previous example.

The vertex gives $A(x)=a(x-4)^2+8$, where $a$ is a fixed nonzero real coefficient. Use $A(0)=0$ at the left endpoint to determine it:

$$\begin{aligned} 0&=16a+8,\\ a&=-\frac12,\\ A(x)&=-\frac12(x-4)^2+8\\ &=-\frac12x^2+4x. \end{aligned}$$

The arch’s domain is $0\leq x\leq8$. The segment’s slope is $(5-2)/(3-0)=1$. Its height at $x=0$ is 2, so

$$B(x)=x+2,\qquad0\leq x\leq3.$$

These domains include the endpoints. The same algebraic rules can be extended to every real $x$, but those full graphs are not the bounded objects in the question.

Graphical route: plot the endpoints and vertex, draw the smooth arch and the straight segment, then read the coordinates of each shared point. A graphing tool can refine the estimate near $(0.76,2.76)$. Show the segment’s endpoint or restrict its domain; otherwise the tool may also report a point on the line’s extension.

2. Shared points have equal heights at the same position

At an intersection, the same pair $(x,y)$ satisfies both rules. Set $A(x)=B(x)$, rather than setting either height to zero:

$$\begin{aligned} -\frac12x^2+4x&=x+2,\\ -\frac12x^2+3x-2&=0,\\ x^2-6x+4&=0,\\ (x-3)^2&=5,\\ x&=3\pm\sqrt5. \end{aligned}$$

In the third line, both sides were multiplied by $-2$; in the fourth, completing the square gives $x^2-6x+9=5$. The two roots locate intersections with the full line. They still need a height and a domain check.

3. Recover both coordinates, then filter the candidates

Use $y=x+2$ to recover each height. Call the point with the smaller $x$-coordinate $P$, and the other point $Q$:

$$\begin{aligned} P&=(3-\sqrt5,\,5-\sqrt5)\\ &\approx(0.764,\,2.764),\\ Q&=(3+\sqrt5,\,5+\sqrt5)\\ &\approx(5.236,\,7.236). \end{aligned}$$

Substitute each exact input into both original rules, using the same sign throughout. Both heights agree:

$$\begin{aligned} A(3\pm\sqrt5)&=5\pm\sqrt5,\\ B(3\pm\sqrt5)&=5\pm\sqrt5. \end{aligned}$$

Both candidates have $0\leq x\leq8$, so both are on the arch. Only $P$ also has $0\leq x\leq3$. Thus the given segment and arch meet once, at $P$, about 0.764 m to the right of the left endpoint and 2.764 m above the ground. $Q$ is a valid intersection of the full graphs, but lies beyond the segment. Rounding comes after the exact calculation.

Two common errors: $A(x)=0$ finds the arch’s ground contacts, not its intersections with the segment. And reporting only the two roots leaves out the heights and the physical-domain decision.

4. Why there can be two, one or no intersections

Subtracting a nonvertical line’s rule from a quadratic rule leaves a quadratic expression. Set that difference equal to zero and use its discriminant $\Delta$: positive gives two distinct full-graph intersections; zero gives one, with a repeated root; negative gives none over the real numbers. Use the coefficients of this difference equation, not the arch’s equation alone. Restricting the domains can remove candidates, but cannot add new intersections.

For the original difference equation above, the coefficients are 1, −6 and 4, giving $\Delta=20>0$. There are two full-graph intersections, but only one contact with the given segment.

Try three new segments

Keep the same arch and its domain. Each question replaces the original segment with a new one on $0\leq x\leq3$, with both endpoints included. First predict the number of contacts from a sketch or graphing tool. Then check algebraically, give every full coordinate pair and apply both domains. For a graphical course route, give coordinate estimates and the domain decision; use the exact working to check your interpretation.

For these three changed segments, the physical contact counts are two, one and zero. For the original segment, two full-graph intersections led to only one physical contact. Keep the equations, coordinate pairs and allowed domains together in your final answer.

From roots to intervals: solve a quadratic inequality

An equation asks where two expressions are equal. An inequality asks for every input where one is greater or smaller. Finding the roots is often the first step, but those roots alone are usually not the answer.

This optional extension assumes you can find quadratic roots and read a parabola. Review linear inequalities and included endpoints if the symbols are unfamiliar. It supports selected Quebec secondary algebra expectations, with year and option depending on the course, and the quadratic part of Ontario MHF4U polynomial inequalities. It is not an additional MPM2D course requirement. If inequalities are not part of your current topic, continue to the original quadratic practice.

1. Use the same parabola to answer a different question

Earlier in this lesson, we used $q(x)=x^2-6x+5=(x-1)(x-5)$. Here $x$ is a real number and $q(x)$ is its output; neither has a physical unit. Solve

$$q(x)\leq0.$$

The equation $q(x)=0$ gives the boundary inputs 1 and 5. The inequality asks where the graph is on or below the horizontal axis. Check the three intervals separated by the roots:

• When $x<1$, both $x-1$ and $x-5$ are negative. Their product is positive. For example, $q(0)=5$.
• When $1<x<5$, $x-1$ is positive and $x-5$ is negative. Their product is negative. For example, $q(3)=-4$.
• When $x>5$, both factors are positive, so the product is positive. For example, $q(6)=5$.

At each root, the product is zero. Because $\leq$ allows equality, include both roots. The solution is every real input from 1 to 5:

$$1\leq x\leq5,\qquad x\in[1,5].$$

On a number line, use filled dots at 1 and 5 and shade the segment between them. On the parabola, the corresponding part of the curve lies on or below the axis. The answer describes horizontal inputs, not a range of vertical outputs.

Why do intervals work? Each linear factor keeps its sign until its own zero is crossed. Thus the product has a fixed sign between consecutive distinct roots. A test input checks that sign; it is not the only input that works.

Change just one symbol: solve $q(x)<0$, then $q(x)>0$.

2. A method you can reuse

1. Bring the comparison to zero. Subtract the same expression from both sides. If asked to solve $f(x)\geq g(x)$, study $f(x)-g(x)\geq0$.
2. Find the real zeros of that difference. Use factoring, a completed square or the quadratic formula from the earlier section. Keep exact values when possible.
3. Determine signs between the zeros. Use the factors or the position of the parabola relative to the axis. Keep the leading coefficient: a negative multiplier reverses the sign of the product.
4. Select the intervals and handle the roots. Include a zero for $\leq$ or $\geq$, exclude it for $<$ or $>$. For a quadratic, the sign does not change at a repeated root.
5. Apply the stated domain and check the original comparison. Test an allowed input and an excluded input when such inputs exist, and check any finite boundary. In a context, some algebraic solutions may not describe a possible situation.

Do not divide an inequality by an expression containing $x$ unless you first account for where that expression is positive, negative or zero. You could reverse the comparison incorrectly or discard solutions. Factoring and checking signs avoids that problem.

3. Keep a negative leading coefficient

Worked example: solve $-2x^2+8x-6\geq0$ for real $x$.

$$-2x^2+8x-6=-2(x-1)(x-3).$$

The roots are 1 and 3. Outside these roots, the two linear factors have the same sign, so their product is positive; multiplying by $-2$ makes the whole expression negative. Between the roots, the factors have opposite signs, so their product is negative; multiplying by $-2$ makes the whole expression positive.

We want a positive or zero value, giving $1\leq x\leq3$. The downward-opening parabola is on or above the axis over this interval. Check the original expression: at $x=2$, $-8+16-6=2\geq0$; at $x=0$, the value is $-6$, which fails; at 1 and 3, the value is zero, so both boundaries work.

An equivalent route divides both sides by $-2$ and reverses the sign:

$$(x-1)(x-3)\leq0.$$

Dividing by the known negative constant is safe when you reverse the comparison. Do not replace the inequality with an equation and stop at the two roots: $x=2$ is also a solution.

4. A repeated root, or no roots, changes the decision

A square is never negative. Use this fact rather than assuming that the sign alternates at every root.

For $p(x)=(x-2)^2$, the only root is 2. The output is positive on both sides of 2, and zero at 2. Therefore:

• $p(x)\leq0$ has the single solution $x=2$.
• $p(x)<0$ has no real solution.
• $p(x)\geq0$ holds for every real $x$.
• $p(x)>0$ holds for every real $x$ except 2.

Now consider $r(x)=(x-2)^2+3$. It has no real roots because $r(x)\geq3$. Nevertheless, $r(x)>0$ holds for every real input, while $r(x)\leq0$ has no solution. “No roots” answers the equation question; the requested inequality still decides whether all or none of the inputs work.

5. Return to the garden: meet a target area

The garden above has 20 m of fencing around all four sides. Its width is $w$ metres, its length is $10-w$ metres, and its area has numerical value $A(w)=w(10-w)$ in square metres. The physical domain is $0<w<10$. What widths give an area of at least 21 m²?

$$\begin{aligned}w(10-w)&\geq21\\-w^2+10w-21&\geq0\\-(w-3)(w-7)&\geq0.\end{aligned}$$

The boundary widths are 3 m and 7 m. The expression is nonnegative between them, including both boundaries, so $3\leq w\leq7$. Every width in that interval is already in the physical domain. At $w=3$ or 7, the area is exactly 21 m²; at $w=5$, it is 25 m²; at $w=2$, it is only 16 m². These checks confirm the endpoints, an interior value and a rejected value.

The question asks for all widths that meet a target. The maximum-area width of 5 m is one valid choice, but it is not the whole answer.

Practise from a sign choice to a new model

Work over the real numbers unless a question gives a narrower domain. For each interval answer, explain a sign and check whether its finite endpoints belong. Try the question before opening its hint or answer.

1. Start with the factors

Solve $(x+2)(x-4)>0$.

2. Change the leading sign

Solve $-x^2+2x+8\geq0$.

3. Move the comparison before finding boundaries

Solve $x^2-4x\leq5$.

4. Do not force a sign change

Solve $2(x-3)^2\leq0$. How do the answers change if $\leq$ is replaced by $<$ or $>$?

5. No real roots does not settle an inequality

Solve $-(x+1)^2-2\geq0$, then solve $-(x+1)^2-2<0$.

6. Keep exact boundaries when factoring is awkward

Solve $2x^2-4x+1\geq0$. Use the earlier quadratic-formula work.

7. Transfer to a different rectangle

A rectangle has perimeter 24 cm. Which widths give an area greater than 32 cm²? State the physical domain and explain whether the boundary widths belong.

8. Diagnose an unsafe division

A student solves $x(x-3)>0$ by dividing both sides by $x$, then writing $x>3$. What is missing? Give the complete solution.

Check your method: Can you explain why a whole interval works, why each endpoint is included or excluded, and how the domain changes the answer? If finding the boundaries was difficult, return to completing the square and the quadratic formula. If the boundary values were correct but the interval was wrong, repeat question 1 and change only the inequality symbol.

6. Independent practice

Try each question before opening its solution.

Choose your next step

If expansion was difficult, revisit brackets and factoring. If your algebra worked but the graph did not, practise labelling the vertex and both kinds of intercept before sketching. If the garden model was difficult, name the two dimensions and write the perimeter equation before the area formula.

Review basic algebra and equations

Tutoring next step

Ask a tutor to work through the bracket and sign steps with you, connect the completed square to the graph, and guide you through a similar question. Tell us your level or course code and the topic you’d like help with. We can confirm topic fit and availability before booking. Tutoring sessions are currently in English.

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Curriculum references and scope

This lesson covers selected quadratic-relations expectations, not an entire course. Quebec year and option requirements vary.

• Ontario MPM2D, 2005, pp. 47–48
• Ontario MPM2D glossary
• Quebec — progression of learning
• Ontario MCR3U and MCF3M, Grades 11–12 curriculum, 2007