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
When a basketball is thrown through the air or bounces on the floor, it's path is a curve called a parabola. Parabolas are modeled with quadratic functions.
A quadratic function is in the form y = ax2 + bx + c where a ≠ 0. The simplest quadratic function is y = x2, so the x2 as the highest exponent of x is what determines that a function is quadratic.
The graph of a quadratic function is shaped like a ∪ and is called a parabola. The lowest or highest point on the parabola is called the vertex. The line through the vertex that goes midway between each side of the parabola is the axis of symmetry and the parabola is symmetric on each side of the axis of symmetry.
Transformations can be applied to quadratic functions like they can to any function. Remember that anything applied inside the function directly to the x will produce a horizontal transformation, and anything applied outside the function will produce a vertical transformation.
y = f(x − h) + k
y = −f(x)
y = f(−x)
y = a·f(bx)
A horizontal stretch in all quadratic functions can be rearranged to describe a vertical shrink because if a graph is stretched wider, then it also looks shorter. This relationship can also be shown algebraically. Thus, the transformations for quadratic functions are usually described as
f(x) = a(x − h)2 + k
This is the standard form of a quadratic function and is sometimes called vertex form.
Describe the transformations to get g(x) = (x − 3)2 + 1 from f(x) = x2.
Solution
Compare g(x) to the standard form to identify a, h, and k.
f(x) = a(x − h)2 + k
g(x) = (x − 3)2 + 1
Inspection shows that a = 1, h = 3, and k = 1.
Because a = 1, there is no stretch or shrink. Because h = 3, the graph is translated 3 right. Because k = 1, is translated 1 up.
Describe the transformations to get \(g\left(x\right)=-\frac{1}{2}\left(x+1\right)^2-2\) from f(x) = x2.
Solution
Compare g(x) to the standard form to identify a, h, and k.
f(x) = a(x − h)2 + k
$$ g\left(x\right)=-\frac{1}{2}\left(x+1\right)^2-2 $$
Inspection shows that \(a = -\frac{1}{2}\), h = −1, and k = −2.
a being negative means the graph is reflected over the x-axis. \(a=\frac{1}{2}\) indicates a vertical shrink by a factor of \(\frac{1}{2}\).
Because h = −1, the graph is translated 1 left. Because k = −2, is translated 2 down.
Write the quadratic function that is transformed from y = x2 by translating 2 unit right and 3 units down with a vertical stretch by a factor of 2.
Solution
Translating right 2 means h = 2. Translating down 3 means k = −3. The vertical stretch by a factor of 2 means a = 2. Substitute those values into the standard form of a quadratic function.
f(x) = a(x − h)2 + k
f(x) = 2(x − 2)2 − 3
The vertex of y = x2 is (0, 0) translating it h units right and k units up means the vertex is now
(0 + h, 0 + k) = (h, k)
That is why standard form of a parabola is also called vertex form, because the vertex is (h, k).
To graph quadratic equations, the simplest way, start by finding the vertex. In standard form, the vertex is (h, k). Then make a table by picking x-values on either side of the vertex and calculating the corresponding y-values. Usually at least a total of 5 or more points is needed to make a nice graph. Plot the points and draw the parabola through the points.
The a in the quadratic function influences the shape of the parabola. If |a| > 1, then the graph is stretched vertically but looks narrower. If 0 < |a| < 1, then the graph is shrunk vertically but looks wider.
a also determines the direction the parabola opens. If a > 0, the parabola opens up. If a < 0, then the parabola opens down.
f(x) = a(x − h)2 + k
Graph f(x) = x2 − 2.
Solution
Compare this to standard form to find a, h, and k to get the vertex.
f(x) = a(x − h)2 + k
f(x) = x2 − 2
This is the same as
f(x) = (x − 0)2 − 2
So a = 1, h = 0, and k = −2. Thus, the vertex (h, k) is (0, −2). Since a is positive, the parabola opens up.
Make a table by picking x-values on either side of the vertex.
| x | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y | 7 | 2 | −1 | −2 | −1 | 2 | 7 |
Plot the points. And draw the parabola through the points.
Graph \(g\left(x\right)=-\frac{1}{2}\left(x+2\right)^2\).
Solution
Compare this to standard form to find a, h, and k to get the vertex.
f(x) = a(x − h)2 + k
$$ f\left(x\right)=-\frac{1}{2}\left(x+2\right)^2 $$
This is the same as
$$ f\left(x\right)=-\frac{1}{2}\left(x+2\right)^2+0 $$
So \(a=-\frac{1}{2}\), h = −2, and k = 0. Thus, the vertex (h, k) is (−2, 0). Since a is negative, the parabola opens down.
Make a table by picking x-values on either side of the vertex.
| x | −5 | −4 | −3 | −2 | −1 | 0 | 1 |
|---|---|---|---|---|---|---|---|
| y | −4.5 | −2 | −0.5 | 0 | −0.5 | −2 | −4.5 |
Plot the points, and draw the parabola through the points.
To find a quadratic model given vertex and another point,
Write the quadratic function for the graph.
Solution
The vertex is (h, k), so from the graph h = −1 and k = −3. Substitute these values into the standard form.
f(x) = a(x − h)2 + k
f(x) = a(x − (−1))2 + (−3)
f(x) = a(x + 1)2 − 3
To find a, find a point on the graph other than the vertex and substitute for x and f(x) such as (0, −1). Then solve for a.
−1 = a(0 + 1)2 − 3
2 = a
This makes sense because the parabola opens up when a > 0 and is narrower when |a| > 1. Substitute that back into the previous function to get the final answer.
f(x) = 2(x + 1)2 − 3
50 #1, 5, 13, 25, 27, 29, 30, and 59 #1, 3, 5, 9, 11, and 76 #1, 3, 5, and Mixed Review = 20
Mixed Review