18.4 Range of a Quadratic Graph

The range of a quadratic graph is the set of all possible y-values that the graph can take.

For a quadratic function$$y=ax^2+bx+c$$the range depends mainly on the turning point and whether the parabola opens upwards or downwards.

Look at the reference diagram above. The two graphs show the two possible situations:$$y=x^2-4x+3$$opens upwards and has a minimum value, while$$y=-x^2+4$$opens downwards and has a maximum value.

Concept

The range tells us all the possible outputs of a function.

Since the output values are shown on the vertical $y$-axis,$$\text{Range}=\text{all possible }y\text{-values}$$

This is different from the domain.$$\text{Domain}\rightarrow x\text{-values}$$$$\text{Range}\rightarrow y\text{-values}$$

For a quadratic graph, the most important point for finding the range is the turning point.

If the parabola opens upwards, the turning point gives the minimum y-value.

If the parabola opens downwards, the turning point gives the maximum y-value.

Formula

A quadratic function has the general form$$y=ax^2+bx+c$$where$$a\ne0$$The $x$-coordinate of the turning point can be found using$$x=-\frac{b}{2a}$$Substitute this value of $x$ into the quadratic equation to find the corresponding $y$-coordinate.

Suppose the turning point is$$(h,k)$$If$$a>0$$the parabola opens upwards, so $k$ is the minimum value.

Therefore,$$y\ge k$$If$$a<0$$the parabola opens downwards, so $k$ is the maximum value.

Therefore,$$y\le k$$So remember:$$a>0\quad\Rightarrow\quad \text{minimum value}$$$$a<0\quad\Rightarrow\quad \text{maximum value}$$

Example Math — Opens Upwards

Look at the left-hand graph in the reference diagram.

The equation is$$y=x^2-4x+3$$

We want to find its range.

Step 1: Determine the direction

Comparing with$$y=ax^2+bx+c$$we have$$a=1$$Since$$a>0$$the parabola opens upwards.

Therefore, the graph has a minimum turning point.

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Step 2: Find the $x$-coordinate of the turning point

Use$$x=-\frac{b}{2a}$$

Here,$$a=1$$and$$b=-4$$so$$x=-\frac{-4}{2(1)}$$$$x=\frac{4}{2}$$$$x=2$$The axis of symmetry is therefore$$x=2$$which agrees with the reference graph.

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Step 3: Find the $y$-coordinate

Substitute$$x=2$$into$$y=x^2-4x+3$$giving$$y=2^2-4(2)+3$$$$y=4-8+3$$$$y=-1$$So the turning point is$$(2,-1)$$and the minimum value of $y$ is$$-1$$

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Step 4: State the range

Because the graph opens upwards, it can take $-1$ and every value greater than $-1$.

Therefore,$$y\ge-1$$So the range is$$\text{Range: }y\ge-1$$This agrees with the left-hand graph in the reference image.

Example Math — Opens Downwards

Now look at the right-hand graph.

The equation is$$y=-x^2+4$$Here,$$a=-1$$Since$$a<0$$the parabola opens downwards.

Therefore, its turning point gives the maximum value.

From the reference graph, the turning point is$$(0,4)$$So the greatest possible value of $y$ is$$4$$Therefore,$$y\le4$$Hence,$$\text{Range: }y\le4$$The graph can take values such as$$4,\ 3,\ 0,\ -5,\ -100,\ldots$$but it can never go above$$y=4$$A+ Insight — Why the Turning Point Determines the Range

An A+ student should understand why, not simply memorize the inequality.

Consider$$y=x^2-4x+3$$

Complete the square:$$y=(x-2)^2-1$$For every real value of $x$,$$(x-2)^2\ge0$$Therefore,$$(x-2)^2-1\ge-1$$which means$$y\ge-1$$So $-1$ is mathematically the smallest possible value of $y$.

That is exactly what we see at the turning point$$(2,-1)$$in the graph.

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A+ Insight — The Downward Case

For$$y=-x^2+4$$we know$$x^2\ge0$$Therefore,$$-x^2\le0$$Adding $4$ gives$$-x^2+4\le4$$so$$y\le4$$Again, the algebra agrees perfectly with the graph.

The maximum occurs when$$x=0$$giving the turning point$$(0,4)$$

Completed Square Form and Range

The completed square form of a quadratic is$$y=a(x-h)^2+k$$The turning point is$$(h,k)$$This makes the range much easier to recognize.

For example,$$y=(x-3)^2-5$$has turning point$$(3,-5)$$and opens upwards.

Therefore,$$y\ge-5$$Now consider$$y=-2(x+1)^2+7$$The turning point is$$(-1,7)$$and the parabola opens downwards.

Therefore,$$y\le7$$For many quadratic questions, completed square form is the fastest way to identify the range.

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Domain vs Range

It is important not to confuse these two ideas.

For the left-hand graph,$$y=x^2-4x+3$$the graph continues indefinitely to the left and right.

Therefore, its domain is$$x\in\mathbb{R}$$But the graph never goes below$$y=-1$$so its range is$$y\ge-1$$For the right-hand graph,$$y=-x^2+4$$the domain is again$$x\in\mathbb{R}$$but the graph never goes above $4$.

Therefore,$$y\le4$$A useful memory rule is:$$\text{Domain}\rightarrow\text{read horizontally}$$$$\text{Range}\rightarrow\text{read vertically}$$Exam Tip

When an exam asks you to find the range of a quadratic graph, first locate the turning point.

Use its y-coordinate, not its $x$-coordinate.

Then look at the direction of the parabola.

If it opens upwards,$$y\ge\text{minimum value}$$If it opens downwards,$$y\le\text{maximum value}$$For example, if the turning point is$$(3,-6)$$and the graph opens upwards, the range is$$y\ge-6$$not$$y\ge3$$

Common Mistake

A very common mistake is using the xx-coordinate of the turning point to write the range.

For the graph

$$y=x^2-4x+3$$the turning point is$$(2,-1)$$The range is not$$y\ge2$$because $2$ is the $x$-coordinate.

The relevant value is the $y$-coordinate:$$-1$$Therefore,$$y\ge-1$$Another common mistake is reversing the inequality.

An upward-opening graph has a minimum, so its $y$-values are greater than or equal to that minimum.

A downward-opening graph has a maximum, so its $y$-values are less than or equal to that maximum.

❓ Question Answer

Q1. What is the range of a quadratic graph?

The range is the set of all possible $y$-values produced by the quadratic function.

Q2. Which coordinate of the turning point determines the range?

The y-coordinate.

Q3. What happens when a>0a>0?

The parabola opens upwards and has a minimum value.

Q4. What happens when a<0a<0?

The parabola opens downwards and has a maximum value.

Q5. What is the range of$$y=x^2-4x+3$$The turning point is$$(2,-1)$$so$$y\ge-1$$Q6. What is the range of$$y=-x^2+4$$The turning point is$$(0,4)$$so$$y\le4$$Q7. Do the roots determine the range?

No. The roots show where$$y=0$$The turning point determines the maximum or minimum value and therefore the range.

Quick Summary

The range of a quadratic graph is the set of possible$$y\text{-values}$$For a quadratic$$y=ax^2+bx+c$$find the turning point.

If the turning point is$$(h,k)$$then:$$a>0\quad\Rightarrow\quad y\ge k$$and$$a<0\quad\Rightarrow\quad y\le k$$From the reference diagram:$$y=x^2-4x+3$$has turning point$$(2,-1)$$so$$\text{Range: }y\ge-1$$while$$y=-x^2+4$$has turning point$$(0,4)$$so$$\text{Range: }y\le4$$The key idea is:$$\text{Upwards}\rightarrow\text{minimum}$$$$\text{Downwards}\rightarrow\text{maximum}$$and the y-coordinate of that turning point gives the boundary of the range.

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Continue with Domain of a Quadratic Graph, Turning Point, Axis of Symmetry, Maximum and Minimum Values, Completing the Square, and Solving Quadratic Equations Graphically.