02.06 Quadratic Graphs

In this lesson you will learn how to:

  • Determine which way a quadratic graph curves
  • Sketch a quadratic graph from its equation, identifying its \(y\)-intercept, roots (if any) and turning point
  • Determine the equation of a quadratic graph if given its roots + a 3rd point
  • Determine the equation of a quadratic graph if given its turning point + a 2nd point
  • Determine the line of symmetry of a quadratic graph

Prerequisites from previous chapters:

  • Expanding Brackets

VIDEO LESSON

INTERACTIVE SELF-STUDY

Consider a quadratic function \(f(x)=ax^2+bx+c\). The graph of \(y=f(x)\) (a quadratic graph) has a curved shape called a parabola, shown adjacent.
You can sketch a quadratic graph by identifying key features.

To determine whether it has a ”\(∪\)” shape or “\(∩\)” shape, we must look at the coefficient of \(x^2\).
Let’s investigate how we can distinguish the cases.

If we look at the two graphs, the first graph goes upwards as you move to the extreme right, while the second goes downwards as you move to the extreme right.
Moving to the right corresponds to \(x\) getting really big and positive (\(x→+∞\)).
Moving upwards corresponds to \(y\) becoming more positive, while moving downwards corresponds to \(y\) becoming more negative.

But if \(x\) gets really big and positive (as \(x→+∞\)), then the biggest term in \(f(x)=ax^2+bx+c\) will be \(ax^2\).
Let’s consider the function \(f(x)=2x^2+100x+500\) to demonstrate this

So the value and hence the sign (positive/negative) of \(y=f(x)\) will be dominated by the contribution from \(ax^2\).

i. Let \(a>0\). If \(x\) gets really big and positive, then what will the sign of \(ax^2\) and hence \(y=f(x)\) be?

ii. Let \(a<0\). If \(x\) gets really big and positive, then what will the sign of \(ax^2\) and hence \(y=f(x)\) be?

Hence the graph of \(f(x)=ax^2+bx+c\) has a:

  • “\(∪\)” shape when \(a>0\)
  • “\(∩\)” shape when \(a<0\)
\(f(x)=ax^2+bx+c\)
with \(a>0\)

\(f(x)=ax^2+bx+c\)
with \(a<0\)

Now that we know which shape it has, we can identify where the graph crosses the coordinate axes.
i. When the graph crosses the \(y\)-axis, what is the value of \(x\)?

ii. Hence what is the value of \(f(x)\) when it crosses the \(y\)-axis? In other words, what is its \(y\)-intercept?

  • The graph of \(f(x)=ax^2+bx+c\) has a \(y\)-intercept of \(c\)

i. When the graph crosses the \(x\)-axis, what is the value of \(y\)?

ii. Hence, what equation should you solve when trying to find the \(x\)-intercepts of the graph of \(y=f(x)\)?
iii. What do you call the solutions to this equation?

  • The graph of \(f(x)=ax^2+bx+c\) has \(x\)-intercepts that are the roots of the equation \(f(x)=0\)

iv. Using the quadratic formula, write down the two roots of the equation \(f(x)=ax^2+bx+c=0\) in terms of \(a, b\) and \(c\). Call one root \(α\) and the other root \(β\)

Example 1 (Walkthrough) – (GRADE D)
We would like to sketch the graph of \(y=2x^2−4x−6\), showing clearly the points where the graph crosses the coordinate axes
a. Write down the \(y\)-intercept and mark it on the axes

b. Find the root(s) by factorising or quadratic formula and mark it on the axes.
c. Is this a positive or negative quadratic
d. Hence sketch the graph

  • If the two roots are equal, that is, if we have a double root at \(x=α\), then the graph will touch the \(x\)-axis at \(x=α\)

Example 2 (Walkthrough) – (GRADE D)
We would like to sketch the graph of \(y=-x^2-4x-4\), showing clearly the points where the graph crosses the coordinate axes
a. Write down the \(y\)-intercept and mark it on the axes

b. Find the root(s) by factorising or quadratic formula and mark it on the axes.
c. Is this a positive or negative quadratic
d. Hence sketch the graph

FINDING EQUATION OF QUADRATIC GRAPH FROM ROOTS

  • If you know that a quadratic function has roots \(α\) and \(β\), then you can immediately write down the factorised form of the quadratic: \(f(x)=a(x−α)(x−β)\)

You must include the \(a\) in front of the factors because there are multiple (infinitely many) quadratic functions that have the same pair of roots, as demonstrated adjacent, and we don’t know which one it is.
Drag the slider tool in the animation adjacent to see how changing the value of \(a\) produces different graphs with the same roots

If we don’t include \(a\), then we may wrongly assume that the coefficient of \(x^2\) is \(1\)

To determine the value of \(a\) and hence the original quadratic function you need to know a third point on the graph.
We can substitute the coordinates of the third point into \(y=a(x−α)(x−β)\) and solve for \(a\)

Example 3 (Walkthrough) – (GRADE C)

We would like to find the equation of the graph adjacent, giving the answer in the form \(y=ax^2+bx+c \)
a. Write down the roots of the function, \(α\) and \(β\), and hence write the quadratic in factorised form: \(f(x)=a(x−α)(x−β)\)

b. Solve for \(a\) by substituting the coordinates of another point on the graph into your factorised form
c. Expand the factorised form

Example 4 – (GRADE C)

Find the equation of the graph adjacent, giving the answer in the form \(y=ax^2+bx+c\)
a. Write down the roots of the function, \(α\) and \(β\), and hence write the quadratic in factorised form: \(y=a(x−α)(x−β)\).

b. Solve for \(a\) by substituting the coordinates of another point on the graph into your factorised form
c. Expand the factorised form

Note that not all quadratic functions have roots as their graphs may lie completely above or completely below the \(x\)-axis. We will discuss the conditions under which this occurs in the next lesson

FINDING TURNING POINT OF QUADRATIC GRAPH FROM EQUATION

We can find the coordinates of the turning point or vertex of the quadratic graph. This is the maximum/minimum point (depending on the shape). We can do this in two ways

For the first method, we can complete the square for the quadratic.

Suppose \(f(x)=a(x-p)^2+q\) with \(a>0\).
i. Write down the minimum value of \(f(x)\)

ii. For what value of \(x\) is this minimum value achieved?

  • The graph of \(f(x)=ax^2+bx+c=a(x-p)^2+q\) has turning point \((p,q)\).

Example 5 (Walkthrough) – (GRADE D)
We would like to sketch the graph of \(y=−2x^2−4x−7\), showing clearly the points where the graph crosses the coordinate axes and the coordinates of the turning point
a. Write down the \(y\)-intercept and mark it on the axes

b. Does the quadratic have roots? If so, find them and mark them on the axes
c. By completing the square, find the coordinates of the turning point and mark it on the axes
d. Is this a positive or negative quadratic?
e. Hence sketch the graph

FINDING EQUATION OF QUADRATIC GRAPH FROM TURNING POINT

  • If you know the turning point of a quadratic is \((p,q)\), then you can immediately write down the completed square form of the quadratic: \(f(x)=a(x−p)^2+q\)

Again you must include the \(a\) in front of the completed square because there are multiple (infinitely many) quadratic functions that have the same turning point, as demonstrated adjacent, and we don’t know which one it is.
Drag the slider tool in the animation adjacent to see how changing the value of \(a\) produces different graphs with the same turning point

If we don’t include \(a\), then we may wrongly assume that the coefficient of \(x^2\) is \(1\)

To determine the value of \(a\) and hence the original quadratic function you need to know a second point on the graph.
We can substitute the coordinates of the second point into \(y=a(x−p)^2+q\) and solve for \(a\)

Example 6 – (GRADE C)
Find the equation of the graph adjacent, giving the answer in the form \(y=ax^2+bx+c\).
a. Write down the coordinates of the turning point \((p, q)\) and hence write the quadratic in completed square form: \(y=a(x−p)^2+q\)

b. Solve for \(a\) by substituting the coordinates of another point on the graph into your completed square form
c. Expand the completed square form

FINDING TURNING POINT OF QUADRATIC GRAPH FROM ROOTS

For the second method of finding the turning point, we should notice that the graph of a quadratic function is symmetric in the vertical line passing through its turning point.

Therefore if a quadratic equation has roots, then the roots must be the same distance away from its turning point.

  • Hence the \(x\)-coordinate of the turning point is halfway in between the roots.
    In other words, it is the average of the \(x\)-coordinates of the roots
    Once we have found the \(x\)-coordinate of the turning point we can substitute it back into the equation to find its \(y\)-coordinate

We can verify this symmetry easily.

Recall that \(f(x)=ax^2+bx+c\) has completed square form \(\displaystyle f(x)=a\left(x+\frac{b}{2a}\right)^2+\left(c−\frac{b^2}{4a}\right)\)
Recall also from the quadratic formula, the two roots of the equation \(f(x)=0\) are:
\(\displaystyle α= \frac{-b-\sqrt{b^2-4ac}}{2a}\) and \(\displaystyle β= \frac{-b+\sqrt{b^2-4ac}}{2a}\)

i. Using the completed square form, write down the coordinates of the turning point of \(f(x)=ax^2+bx+c\)

ii. Verify that the average of \(α\) and \(β\) is the \(x\)-coordinate of the turning point.

  • Hence the graph of a quadratic function \(f(x)=ax^2+bx+c\), has a line of symmetry \(\displaystyle x=−\frac{b}{2a}\)
  • Only quadratic graphs have this general symmetry between the roots and coordinates of the turning point.
    This is generally not true for other polynomial graphs (cubics, quartics, etc.)
Example 7 (Walkthrough) – (GRADE D)
We would like to sketch the graph of \(y=x^2−2x−8\), showing clearly the points where the graph crosses the coordinate axes, the coordinates of the turning point.
a. Write down the \(y\)-intercept and mark it on the axes
b. Find the roots by factorising or using the quadratic formula and mark it on the axes
c. Find the coordinates of the turning point, either by completing the square or using the roots and equation, and mark it on the diagram
d. Is this a positive or negative quadratic?
e. Hence sketch the graph
f. State the equation of the line of symmetry of the graph

END OF LESSON!

You can now answer:
EDEXCEL Pure Year 1 Ex 2F – All Questions
OCR A Student Book 1 Ex 3C – Q3

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