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

Contents

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

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

- \(y_1 = x^2\)
- \(y_2 = (x – 2)^2\)
- \(y_3 = (x – 1)^2\)
- \(y_4 = (x + 1)^2\)
- \(y_5 = (x + 2)^2\)

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 \(p\)**On the same system of axes, plot the following graphs:

- \(y_1 = x^2 + 2\)
- \(y_2 = (x – 2)^2 – 1\)
- \(y_3 = (x – 1)^2 + 1\)
- \(y_4 = (x + 1)^2 + 1\)
- \(y_5 = (x + 2)^2 – 1\)

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 \(q\)**Consider the three functions given below and answer the questions that follow:

- \(y_1 = (x – 2)^2 + 1\)
- \(y_2 = 2(x – 2)^2 + 1\)
- \(y_3 = -\cfrac{1}{2}(x – 2)^2 + 1\)

What is the value of \(a\) for \(y_2\)?

Does \(y_1\) have a minimum or maximum turning point?

What are the coordinates of the turning point of \(y_2\)?

Compare the graphs of \(y_1\) and \(y_2\). Discuss the similarities and differences.

What is the value of \(a\) for \(y_3\)?

Will the graph of \(y_3\) be narrower or wider than the graph of \(y_1\)?

Determine the coordinates of the turning point of \(y_3\).

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 \((-p;q)\).

**Alternative form for quadratic equations:**

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

- Show that the \(x\)-value for the turning point of \(h(x) = ax^2 + bx + c\) is given by \(x = -\cfrac{b}{2a}\).
- 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.

### Plot the points and sketch the graph

### 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 \((7;0)\).

### Plot the points and sketch the graph

### 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.

Carl’s answer:

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

Eric’s answer:

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 \(y = x^2 – 2x -3\).

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

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

### Determine the new equation of the shifted parabola

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.

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 \(y = a(x – x_1)(x – x_2)\).

| \(x\)-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 \(y = a(x -x_1)(x – x_2)\).

| \(x\)-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 \(y = a(x + p)^2 + q\).

| 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: \(y = \cfrac{1}{4}(x + 3)^2 + 1\) |