Algebra 2 by Richard Wright
Are you not my student and
has this helped you?
Objectives:
SDA NAD Content Standards (2018): AII.5.1, AII.5.3, AII.7.1
In the previous lesson, it was given that standard form of a quadratic function is
y = a(x – h)2 + k
where (h, k) is the vertex and x = h is the axis of symmetry. This lesson introduces two other forms of a quadratic function.
Intercept form is
y = a(x – p)(x – q)
where p and q are the x-intercepts.
Because of symmetry, the axis of symmetry is halfway between the x-intercepts.
$$ x=\frac{p+q}{2} $$
The vertex is on the axis of symmetry, so it can be found by substituting the x-coordinate of the axis of symmetry into the original function to find the y-value.
Both the standard form and vertex form of a quadratic function can be simplified by multiplying out the expression. This will produce the general form which is
y = ax2 + bx + c
The axis of symmetry is
$$ x=-\frac{b}{2a} $$
and the vertex is found by substituting the x-value of the axis of symmetry into the function to get the y-value.
$$ \left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right) $$
The table compares these three forms of quadratic functions.
| Form | Function | Axis of Symmetry | Vertex | a | Special Properties |
|---|---|---|---|---|---|
| Standard (Vertex) | y = a(x – h)2 + k | x = h | (h, k) | a is vertical stretch factor | vertex (h, k) |
| Intercept | y = a(x – p)(x – q) | \(x=\frac{p+q}{2}\) | \(\left(\frac{p+q}{2},\ f\left(\frac{p+q}{2}\right)\right)\) | a is vertical stretch factor | p and q are x-intercepts |
| General | y = ax2 + bx + c | \(x=-\frac{b}{2a}\) | \(\left(-\frac{b}{2a},\ f\left(-\frac{b}{2a}\right)\right)\) | a is vertical stretch factor | none |
To graph a quadratic function,
Graph \(y=\frac{1}{2}\left(x-3\right)\left(x+1\right)\) and identify the vertex and axis of symmetry.
Solution
This is intercept form \(y=a\left(x-p\right)\left(x-q\right)\) with \(a=\frac{1}{2}\) and p = 3 and q = −1. The axis of symmetry is
$$ x=\frac{p+q}{2} $$
$$ x=\frac{3+\left(-1\right)}{2}\rightarrow x=\mathbf{1} $$
The vertex is found by substituting the axis of symmetry into the function.
$$ y=\frac{1}{2}\left(\mathbf{1}-3\right)\left(\mathbf{1}+1\right)=-\mathbf{2} $$
The vertex is (1, −2).
Make a table of values with points on both sides of the vertex.
| x | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| y | 2.5 | 0 | −1.5 | −2 | −1.5 | 0 | 2.5 |
Graph the points and draw the curve through the points.
Graph y = x2 + 2x – 3 and identify the vertex and axis of symmetry.
Solution
This is general form y = ax2 + bx + c with a = 1, b = 2, and c = −3. The axis of symmetry is
$$ x=-\frac{b}{2a} $$
$$ x=-\frac{2}{2\left(1\right)}=-\mathbf{1} $$
The vertex is found by substituting the axis of symmetry into the function.
y = (–1)2 + 2(–1) – 3= –4
The vertex is (−1, −4).
Make a table of values with points on both sides of the vertex.
| x | −4 | −3 | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|---|
| y | 5 | 0 | −3 | −4 | −3 | 0 | 5 |
Graph the points and draw a parabola through the points.
To write a quadratic function in intercept form,
Write the quadratic function whose x-intercepts are 2 and −5 and passes through (1, −12).
Solution
The x-intercepts are 2 and −5, so p = 2 and q = −5. Substitute these into intercept form.
y = a(x – p)(x – q)
y = a(x – 2)(x + 5)
Substitute the other point for (x, y).
–12 = a(1 – 2)(1 + 5)
–12 = –6a
2 = a
Write the function by substituting a, p, and q into intercept form.
y = 2(x – 2)(x + 5)
Write the quadratic function given in the graph.
Solution
The x-intercepts are −4 and 0, so p = −4 and q = 0. Substitute these into intercept form.
y = a(x – p)(x – q)
y = a(x + 4)(x – 0)
x – 0 = x, so this simplifies to become
y = ax(x + 4)
Another point on the graph is (−2, 2). Substitute that point for (x, y).
2 = a(–2)(–2 + 4)
2 = –4a
$$ -\frac{1}{2}=a $$
Write the function by substituting a, p, and q into intercept form.
$$ y=-\frac{1}{2}x\left(x+4\right) $$
Note: This also could have been done using vertex form.
A football player kicks a ball off the ground. The path of the ball can be modeled by \(y=-\frac{1}{20}x(x-90)\) where x and y are in feet. How far did the ball go? How high did the ball go?
Solution
This function is in intercept form, y = a(x – p)(x – q), with \(a=-\frac{1}{20}\), p = 0, and q = 90. The ball is on the ground at the x-intercepts which are 0 and 90. Thus, the ball went 90 feet.
The highest point is at the vertex. In intercept form, the vertex is
$$ x=\frac{p+q}{2} $$
$$ x=\frac{0+90}{2}=45 $$
$$ y=-\frac{1}{20}\left(45\right)\left(45-90\right)=101.25 $$
Height is in the vertical, or y, direction. The ball went 101.25 feet up.
59 #17, 19, 21, 23, 29, 45, 47, 49, 50, 65, and 76 #7, 9, 11, 15, 17, and Mixed Review = 20
Mixed Review