Mathematics » Functions II » Quadratic Functions

# Functions of the Form y = a(x + p)2 + q

## Functions of the form $$y=a{(x+p)}^{2}+q$$

We now consider parabolic functions of the form $$y=a{(x+p)}^{2}+q$$ and the effects of parameter $$p$$.

## Optional Investigation: The effects of $$a$$, $$p$$ and $$q$$ on a parabolic graph

1. On the same system of axes, plot the following graphs:

1. $$y_1 = x^2$$
2. $$y_2 = (x – 2)^2$$
3. $$y_3 = (x – 1)^2$$
4. $$y_4 = (x + 1)^2$$
5. $$y_5 = (x + 2)^2$$

Use your sketches of the functions above to complete the following table:

 $$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ $$y_5$$ $$x$$-intercept(s) $$y$$-intercept turning point axis of symmetry domain range effect of $$p$$
2. On the same system of axes, plot the following graphs:

1. $$y_1 = x^2 + 2$$
2. $$y_2 = (x – 2)^2 – 1$$
3. $$y_3 = (x – 1)^2 + 1$$
4. $$y_4 = (x + 1)^2 + 1$$
5. $$y_5 = (x + 2)^2 – 1$$

Use your sketches of the functions above to complete the following table:

 $$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ $$y_5$$ $$x$$-intercept(s) $$y$$-intercept turning point axis of symmetry domain range effect of $$q$$
3. Consider the three functions given below and answer the questions that follow:

• $$y_1 = (x – 2)^2 + 1$$
• $$y_2 = 2(x – 2)^2 + 1$$
• $$y_3 = -\cfrac{1}{2}(x – 2)^2 + 1$$
1. What is the value of $$a$$ for $$y_2$$?

2. Does $$y_1$$ have a minimum or maximum turning point?

3. What are the coordinates of the turning point of $$y_2$$?

4. Compare the graphs of $$y_1$$ and $$y_2$$. Discuss the similarities and differences.

5. What is the value of $$a$$ for $$y_3$$?

6. Will the graph of $$y_3$$ be narrower or wider than the graph of $$y_1$$?

7. Determine the coordinates of the turning point of $$y_3$$.

8. Compare the graphs of $$y_1$$ and $$y_3$$. Describe any differences.

### The effect of the parameters on $$y = a(x + p)^2 + q$$

The effect of $$p$$ is a horizontal shift because all points are moved the same distance in the same direction (the entire graph slides to the left or to the right).

• For $$p>0$$, the graph is shifted to the left by $$p$$ units.

• For $$p<0$$, the graph is shifted to the right by $$p$$ units.

The value of $$p$$ also affects whether the turning point is to the left of the $$y$$-axis $$(p>0)$$ or to the right of the $$y$$-axis $$(p<0)$$. The axis of symmetry is the line $$x = -p$$.

The effect of $$q$$ is a vertical shift. The value of $$q$$ affects whether the turning point of the graph is above the $$x$$-axis $$(q>0)$$ or below the $$x$$-axis $$(q<0)$$.

The value of $$a$$ affects the shape of the graph. If $$a<0$$, the graph is a “frown” and has a maximum turning point. If $$a>0$$ then the graph is a “smile” and has a minimum turning point. When $$a = 0$$, the graph is a horizontal line $$y = q$$.

 $$p>0$$ $$p<0$$ $$a<0$$ $$a>0$$ $$a<0$$ $$a>0$$ $$q>0$$ $$q<0$$

### Discovering the characteristics

For functions of the general form $$f(x) = y = a(x + p)^2 + q$$:

### Domain and range

The domain is $$\{x:x\in ℝ\}$$ because there is no value of $$x$$ for which $$f(x)$$ is undefined.

The range of $$f(x)$$ depends on whether the value for $$a$$ is positive or negative. If $$a>0$$ we have: $\begin{array}{r@{\;}c@{\;}l@{\quad}l} (x + p)^2 & \geq & 0 & (\text{perfect square is always positive}) \\ \therefore a(x + p)^2 & \geq & 0 & (a \text{ is positive}) \\ \therefore a(x + p)^2 + q & \geq & q & \\ \therefore f(x) & \geq & q & \end{array}$

The range is therefore $$\{ y: y \geq q, y \in \mathbb{R} \}$$ if $$a > 0$$. Similarly, if $$a < 0$$, the range is $$\{ y: y \leq q, y \in \mathbb{R} \}$$.

## Example

### Question

State the domain and range for $$g(x) = -2(x – 1)^2 + 3$$.

### Determine the domain

The domain is $$\{x: x \in \mathbb{R} \}$$ because there is no value of $$x$$ for which $$g(x)$$ is undefined.

### Determine the range

The range of $$g(x)$$ can be calculated from: \begin{align*} (x – 1)^2& \geq 0 \\ -2(x – 1)^2& \leq 0 \\ -2(x – 1)^2 + 3 & \leq 3 \\ g(x) & \leq 3\end{align*}Therefore the range is $$\{g(x): g(x) \leq 3 \}$$ or in interval notation $$(-\infty; 3]$$.

Notice in the example above that it helps to have the function in the form $$y = a(x + p)^2 + q$$.

We use the method of completing the square to write a quadratic function of the general form $$y = ax^2 + bx +c$$ in the form $$y = a(x + p)^2 + q$$ (see Chapter $$\text{2}$$).

### Intercepts

The $$y$$-intercept:

Every point on the $$y$$-axis has an $$x$$-coordinate of $$\text{0}$$, therefore to calculate the $$y$$-intercept we let $$x=0$$.

For example, the $$y$$-intercept of $$g(x) = (x – 1)^2 + 5$$ is determined by setting $$x=0$$: \begin{align*} g(x) &= (x – 1)^2 + 5 \\ g(0) &= (0 – 1)^2 + 5 \\ &= 6 \end{align*} This gives the point $$(0;6)$$.

The $$x$$-intercept:

Every point on the $$x$$-axis has a $$y$$-coordinate of $$\text{0}$$, therefore to calculate the $$x$$-intercept we let $$y=0$$.

For example, the $$x$$-intercept of $$g(x) = (x – 1)^2 + 5$$ is determined by setting $$y=0$$: \begin{align*} g(x) &= (x – 1)^2 + 5 \\ 0 &= (x – 1)^2 + 5 \\ -5 &= (x – 1)^2 \end{align*} which has no real solutions. Therefore, the graph of $$g(x)$$ lies above the $$x$$-axis and does not have any $$x$$-intercepts.

### Turning point

The turning point of the function $$f(x) = a(x+p)^2 + q$$ is determined by examining the range of the function:

• If $$a > 0$$, $$f(x)$$ has a minimum turning point and the range is $$[q;\infty)$$:

The minimum value of $$f(x)$$ is $$q$$.

If $$f(x) = q$$, then $$a(x+p)^2 = 0$$, and therefore $$x = -p$$.

This gives the turning point $$(-p;q)$$.

• If $$a < 0$$, $$f(x)$$ has a maximum turning point and the range is $$(-\infty;q]$$:

The maximum value of $$f(x)$$ is $$q$$.

If $$f(x) = q$$, then $$a(x+p)^2 = 0$$, and therefore $$x = -p$$.

This gives the turning point $$(-p;q)$$.

Therefore the turning point of the quadratic function $$f(x) = a(x+p)^2 + q$$ is $$(-p;q)$$.

We can also write the quadratic equation in the form

$y = a(x – p)^2 +q$

The effect of $$p$$ is still a horizontal shift, however notice that:

• For $$p>0$$, the graph is shifted to the right by $$p$$ units.

• For $$p<0$$, the graph is shifted to the left by $$p$$ units.

The turning point is $$(p;q)$$ and the axis of symmetry is the line $$x = p$$.

## Example

### Question

Determine the turning point of $$g(x) = 3x^2 – 6x – 1$$.

### Write the equation in the form $$y = a(x+p)^2 + q$$

We use the method of completing the square: \begin{align*} g(x )&= 3x^2 – 6x – 1 \\ &= 3(x^2 – 2x) – 1 \\ &= 3 ( (x-1)^2 – 1 ) -1 \\ &= 3(x-1)^2 – 3 -1 \\ &= 3(x-1)^2 – 4 \end{align*}

### Determine turning point $$(-p;q)$$

From the equation $$g(x) = 3(x-1)^2 – 4$$ we know that the turning point for $$g(x)$$ is $$(1;-4)$$.

## Example

### Question

1. Show that the $$x$$-value for the turning point of $$h(x) = ax^2 + bx + c$$ is given by $$x = -\cfrac{b}{2a}$$.
2. Hence, determine the turning point of $$k(x) = 2 – 10x + 5x^2$$.

### Write the equation in the form $$y = a(x+p)^2 + q$$ and show that $$p = \cfrac{b}{2a}$$

We use the method of completing the square: \begin{align*} h(x)&= ax^2 + bx + c \\ &= a ( x^2 + \cfrac{b}{a}x + \cfrac{c}{a} ) \end{align*}

Take half the coefficient of the $$x$$ term and square it; then add and subtract it from the expression. \begin{align*} h(x) &= a ( x^2 + \cfrac{b}{a}x + ( \cfrac{b}{2a} )^2 – ( \cfrac{b}{2a} )^2 + \cfrac{c}{a} ) \\ &= a ( (x + \cfrac{b}{2a} )^2 – \cfrac{b^2}{4a^2} + \cfrac{c}{a} ) \\ &= a ( (x + \cfrac{b}{2a} )^2 – \cfrac{b^2 -4ac}{4a^2} ) \\ &= a (x + \cfrac{b}{2a} )^2 – \cfrac{b^2 -4ac}{4a} \end{align*} From the above we have that the turning point is at $$x = -p = – \cfrac{b}{2a}$$ and $$y = q = – \cfrac{b^2 -4ac}{4a}$$.

### Determine the turning point of $$k(x)$$

Write the equation in the general form $$y = ax^2 + bx + c$$. $k(x) = 5x^2 -10x + 2$ Therefore $$a = 5$$; $$b = -10$$; $$c = 2$$.

Use the results obtained above to determine $$x = – \cfrac{b}{2a}$$: \begin{align*} x &= -\left(\cfrac{-10}{2(5)}\right) \\ &= 1 \end{align*}

Substitute $$x = 1$$ to obtain the corresponding $$y$$-value: \begin{align*} y &= 5x^2 -10x + 2 \\ &= 5(1)^2 -10(1) + 2\\ &= 5 – \text{10} + 2\\ &= -3 \end{align*} The turning point of $$k(x)$$ is $$(1;-3)$$.

### Sketching graphs of the form $$f(x)=a{(x+p)}^{2}+q$$

In order to sketch graphs of the form $$f(x)=a{(x+p)}^{2}+q$$, we need to determine five characteristics:

• sign of $$a$$

• turning point

• $$y$$-intercept

• $$x$$-intercept(s) (if they exist)

• domain and range

## Example

### Question

Sketch the graph of $$y = -\cfrac{1}{2}(x + 1)^2 – 3$$.

Mark the intercepts, turning point and the axis of symmetry. State the domain and range of the function.

### Examine the equation of the form $$y = a(x + p)^2 + q$$

We notice that $$a < 0$$, therefore the graph is a “frown” and has a maximum turning point.

### Determine the turning point $$(-p;q)$$

From the equation we know that the turning point is $$(-1; -3)$$.

### Determine the axis of symmetry $$x = -p$$

From the equation we know that the axis of symmetry is $$x = -1$$.

### Determine the $$y$$-intercept

The $$y$$-intercept is obtained by letting $$x = 0$$: \begin{align*} y &= -\cfrac{1}{2} ((0) + 1 )^2 – 3\\ &= -\cfrac{1}{2} – 3\\ &= -3\cfrac{1}{2} \end{align*} This gives the point $$(0;-3\cfrac{1}{2})$$.

### Determine the $$x$$-intercepts

The $$x$$-intercepts are obtained by letting $$y = 0$$: \begin{align*} 0 &= -\cfrac{1}{2} (x + 1 )^2 – 3\\ 3 &= -\cfrac{1}{2} (x + 1 )^2 \\ -6 &= (x + 1 )^2 \end{align*} which has no real solutions. Therefore, there are no $$x$$-intercepts and the graph lies below the $$x$$-axis.

### State the domain and range

Domain: $$\{ x: x \in \mathbb{R} \}$$

Range: $$\{ y: y \leq -3, y \in \mathbb{R} \}$$

## Example

### Question

Sketch the graph of $$y = \cfrac{1}{2}x^2 – 4x + \cfrac{7}{2}$$.

Determine the intercepts, turning point and the axis of symmetry. Give the domain and range of the function.

### Examine the equation of the form $$y = ax^2 + bx +c$$

We notice that $$a > 0$$, therefore the graph is a “smile” and has a minimum turning point.

### Determine the turning point and the axis of symmetry

Check that the equation is in standard form and identify the coefficients.

$a = \cfrac{1}{2}; \qquad b = -4; \qquad c = \cfrac{7}{2}$

Calculate the $$x$$-value of the turning point using \begin{align*} x &= -\cfrac{b}{2a} \\ &= -\left(\cfrac{-4}{2( \cfrac{1}{2} )}\right) \\ &= 4 \end{align*} Therefore the axis of symmetry is $$x = 4$$.

Substitute $$x = 4$$ into the original equation to obtain the corresponding $$y$$-value. \begin{align*} y &= \cfrac{1}{2}x^2 – 4x + \cfrac{7}{2} \\ &= \cfrac{1}{2}(4)^2 – 4(4) + \cfrac{7}{2} \\ &= 8 -16 +\cfrac{7}{2} \\ &= -4\cfrac{1}{2} \end{align*} This gives the point $$( 4; -4\cfrac{1}{2} )$$.

### Determine the $$y$$-intercept

The $$y$$-intercept is obtained by letting $$x = 0$$: \begin{align*} y &= \cfrac{1}{2}(0)^2 – 4(0) + \cfrac{7}{2}\\ &= \cfrac{7}{2} \end{align*} This gives the point $$(0;\cfrac{7}{2})$$.

### Determine the $$x$$-intercepts

The $$x$$-intercepts are obtained by letting $$y = 0$$: \begin{align*} 0 &= \cfrac{1}{2}x^2 – 4x + \cfrac{7}{2} \\ &= x^2 – 8x + 7 \\ &= (x – 1)(x – 7) \end{align*} Therefore $$x = 1$$ or $$x = 7$$. This gives the points $$(1;0)$$ and $$(7;0)$$.

### State the domain and range

Domain: $$\{ x: x \in \mathbb{R} \}$$

Range: $$\{ y: y \geq -4\cfrac{1}{2}, y \in \mathbb{R} \}$$

## Optional Investigation: Shifting the equation of a parabola

Carl and Eric are doing their Mathematics homework and decide to check each others answers.

### Homework question:

If the parabola $$y = 3x^2 + 1$$ is shifted $$\text{2}$$ units to the right, determine the equation of the new parabola.

A shift to the right means moving in the positive $$x$$ direction, therefore $$x$$ is replaced with $$x + 2$$ and the new equation is $$y = 3(x + 2)^2 + 1$$.

We replace $$x$$ with $$x – 2$$, therefore the new equation is $$y = 3(x – 2)^2 + 1$$.

Work together in pairs. Discuss the two different answers and decide which one is correct. Use calculations and sketches to help explain your reasoning.

### Writing an equation of a shifted parabola

The parabola is shifted horizontally:

• If the parabola is shifted $$m$$ units to the right, $$x$$ is replaced by $$(x-m)$$.

• If the parabola is shifted $$m$$ units to the left, $$x$$ is replaced by $$(x+m)$$.

The parabola is shifted vertically:

• If the parabola is shifted $$n$$ units down, $$y$$ is replaced by $$(y+n)$$.

• If the parabola is shifted $$n$$ units up, $$y$$ is replaced by $$(y-n)$$.

## Example

### Question

Given $$y = x^2 – 2x -3$$.

1. If the parabola is shifted $$\text{1}$$ unit to the right, determine the new equation of the parabola.

2. If the parabola is shifted $$\text{3}$$ units down, determine the new equation of the parabola.

### Determine the new equation of the shifted parabola

1. The parabola is shifted $$\text{1}$$ unit to the right, so $$x$$ must be replaced by $$(x-1)$$. \begin{align*} y &= x^2 – 2x -3\\ &= (x-1)^2 – 2(x-1) -3 \\ &= x^2 – 2x + 1 -2x + 2 – 3\\ &= x^2 – 4x \end{align*} Be careful not to make a common error: replacing $$x$$ with $$x+1$$ for a shift to the right.

2. The parabola is shifted $$\text{3}$$ units down, so $$y$$ must be replaced by $$(y+3)$$. \begin{align*} y + 3&= x^2 – 2x -3\\ y &= x^2 – 2x -3 – 3 \\ &= x^2 – 2x -6 \end{align*}

### Finding the equation of a parabola from the graph

If the intercepts are given, use $$y = a(x – x_1)(x – x_2)$$.

 Example: $$x$$-intercepts: $$(-1;0)$$ and $$(4;0)$$\begin{aligned} y &= a(x -x_1)(x – x_2) \\ &= a(x + 1)(x – 4) \\ &= ax^2 – 3ax – 4a \end{aligned}$$y$$-intercept: $$(0;2)$$\begin{aligned} -4a &= 2 \\ a &= -\cfrac{1}{2} \end{aligned}Equation of the parabola:\begin{aligned} y &= ax^2 – 3ax – 4a \\ &= -\cfrac{1}{2}x^2 – 3(-\cfrac{1}{2})x – 4(-\cfrac{1}{2}) \\ &= -\cfrac{1}{2}x^2 + \cfrac{3}{2}x + 2 \end{aligned}

If the $$x$$-intercepts and another point are given, use $$y = a(x -x_1)(x – x_2)$$.

 Example: $$x$$-intercepts: $$(1;0)$$ and $$(5;0)$$\begin{aligned} y &= a(x -x_1)(x – x_2) \\ &= a(x – 1)(x – 5) \\ &= ax^2 – 6ax + 5a \end{aligned}Substitute the point: $$(-1;\text{12})$$\begin{aligned} \text{12} &= a(-1)^2 – 6a(-1) + 5a \\ \text{12} &= a + 6a + 5a \\ \text{12} &= 12a \\ 1 &= a \end{aligned}Equation of the parabola:\begin{aligned} y &= ax^2 – 6ax + 5a \\ &= x^2 – 6x + 5 \end{aligned}

If the turning point and another point are given, use $$y = a(x + p)^2 + q$$.

 Example: Turning point: $$(-3;1)$$\begin{aligned} y &= a(x + p)^2 + q \\ &= a(x + 3)^2 + 1 \\ &= ax^2 + 6ax + 9a + 1 \end{aligned}Substitute the point: $$(1;5)$$\begin{aligned} 5 &= a(1)^2 + 6a(1) + 9a + 1 \\ 4 &= 16a \\ \cfrac{1}{4} &= a \end{aligned}Equation of the parabola:$$y = \cfrac{1}{4}(x + 3)^2 + 1$$