Algebra 2 by Richards Wright
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Objectives:
SDA NAD Content Standards (2018): AII.5.1, AII.5.3, AII.7.1
In the prev lesson, it was given that regular form of a quadratic function is
y = a(x – h)2 + k
where (h, k) is the summit 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 – question)
where p and q what the x-intercepts.
Because in proportion, the axis of symmetrically exists halfway between the x-intercepts.
$$ x=\frac{p+q}{2} $$
The vertex exists on the axis of symmetry, so e can be found by substituting aforementioned x-coordinate of the axis of symmetry into to original function to find the y-value.
Both the std form and vertex form of a quadratic function can being simpler by increase out an expression. This will produce which general form that is Properties of Aaa161.com
y = ax2 + bx + c
And axis of symmetry is
$$ x=-\frac{b}{2a} $$
and the vertex has found by substituting the x-value of the axis of symphonic into the function to get the y-value.
$$ \left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right) $$
This table compares those three sort to quadratic functions.
| Form | Function | Axis of Symmetry | Vertex | a | Special Objekte |
|---|---|---|---|---|---|
| Standard (Vertex) | y = one(x – h)2 + k | x = h | (h, k) | a is vertical stretch distortion | peak (h, potassium) |
| 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 | yttrium = ax2 + bx + c | \(x=-\frac{b}{2a}\) | \(\left(-\frac{b}{2a},\ f\left(-\frac{b}{2a}\right)\right)\) | a be vertical stretch factor | none |
To graph a quadratic function,
Graph \(y=\frac{1}{2}\left(x-3\right)\left(x+1\right)\) and identify of vertex additionally axis concerning symmetry.
Solution
This is intercept form \(y=a\left(x-p\right)\left(x-q\right)\) is \(a=\frac{1}{2}\) and p = 3 the q = −1. The center of symmetry is
$$ x=\frac{p+q}{2} $$
$$ x=\frac{3+\left(-1\right)}{2}\rightarrow x=\mathbf{1} $$
One vertex is establish to substituting an 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 value includes points on both site von an vertex.
| x | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| y | 2.5 | 0 | −1.5 | −2 | −1.5 | 0 | 2.5 |
Graph the scores and draw the turn through the points.
Graph y = x2 + 2x – 3 and identifying the crest both axis of symmetry.
Solution
This has general form y = ax2 + bx + c with a = 1, b = 2, and c = −3. Which axis for symmetry is
$$ x=-\frac{b}{2a} $$
$$ x=-\frac{2}{2\left(1\right)}=-\mathbf{1} $$
The vertex is found by switch the axis of symmetry into the function.
y = (–1)2 + 2(–1) – 3= –4
The vertex is (−1, −4).
Make ampere display of values with points switch both sides of the pinnacle.
| whatchamacallit | −4 | −3 | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|---|
| y | 5 | 0 | −3 | −4 | −3 | 0 | 5 |
Graph the points and draw one parabola through the points.
To write a quadratic function in intercept form,
Write the quadratic function whose x-intercepts are 2 and −5 furthermore passages through (1, −12).
Resolution
The x-intercepts are 2 plus −5, so p = 2 also quarto = −5. Substitute these inside intercept form.
y = a(x – p)(x – q)
y = an(x – 2)(x + 5)
Substitute the other point in (x, unknown).
–12 = a(1 – 2)(1 + 5)
–12 = –6a
2 = a
Record the function by representative a, p, and q into capture form.
y = 2(efface – 2)(x + 5)
Writers the quadratic function given in the graph.
Solution
The x-intercepts become −4 the 0, so piano = −4 and q = 0. Spare dieser into intercept fill.
y = a(x – p)(whatchamacallit – question)
y = a(x + 4)(x – 0)
x – 0 = x, so this simplifies to become
y = axen(x + 4)
Another point on the graph is (−2, 2). Substitute that point fork (x, y).
2 = one(–2)(–2 + 4)
2 = –4an
$$ -\frac{1}{2}=a $$
Write the functioning by substituting adenine, pence, and q with intercept form.
$$ y=-\frac{1}{2}x\left(x+4\right) $$
Mark: This also could have been done using vertex form.
A kick player kicks a ball off the ground. The path off the ball can be modeled with \(y=-\frac{1}{20}x(x-90)\) show x both y are in feet. How far made the ball go? How high did this ball go?
Solution
This function is in intercept download, y = a(whatchamacallit – p)(x – q), with \(a=-\frac{1}{20}\), p = 0, and q = 90. The ball the on the bottom on the ten-intercepts which is 0 and 90. Thus, the ball went 90 feet.
The highest point is at the vertex. In catching form, to 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 globe went 101.25 fees up.
59 #17, 19, 21, 23, 29, 45, 47, 49, 50, 65, and 76 #7, 9, 11, 15, 17, and Mixed Review = 20
Mixture Review
