1. First we are going to find the vertex of the quadratic function [tex]f(x)=2x^2+8x+1[/tex]. To do it, we are going to use the vertex formula. For a quadratic function of the form [tex]f(x)=ax^2+bx +c[/tex], its vertex [tex](h,k)[/tex] is given by the formula [tex]h= \frac{-b}{2a} [/tex]; [tex]k=f(h)[/tex].
We can infer from our problem that [tex]a=2[/tex] and [tex]b=8[/tex], sol lets replace the values in our formula:
[tex]h= \frac{-8}{2(2)} [/tex]
[tex]h= \frac{-8}{4} [/tex]
[tex]h=-2[/tex]
Now, to find [tex]k[/tex], we are going to evaluate the function at [tex]h[/tex]. In other words, we are going to replace [tex]x[/tex] with -2 in the function:
[tex]k=f(-2)=2(-2)^2+8(-2)+1[/tex]
[tex]k=f(-2)=2(4)-16+1[/tex]
[tex]k=f(-2)=8-16+1[/tex]
[tex]k=f(-2)=-7[/tex]
[tex]k=-7[/tex]
So, our first point, the vertex [tex](h,k)[/tex] of the parabola, is the point [tex](-2,-7)[/tex].
To find our second point, we are going to find the y-intercept of the parabola. To do it we are going to evaluate the function at zero; in other words, we are going to replace [tex]x[/tex] with 0:
[tex]f(x)=2x^2+8x+1[/tex]
[tex]f(0)=2(0)^2+(0)x+1[/tex]
[tex]f(0)=1[/tex]
So, our second point, the y-intercept of the parabola, is the point (0,1)
We can conclude that using the vertex (-2,-7) and a second point we can graph [tex]f(x)=2x^2+8x+1[/tex] as shown in picture 1.
2. The vertex form of a quadratic function is given by the formula: [tex]f(x)=a(x-h)^2+k[/tex]
where
[tex](h,k)[/tex] is the vertex of the parabola.
We know from our previous point how to find the vertex of a parabola. [tex]h= \frac{-b}{2a} [/tex] and [tex]k=f(h)[/tex], so lets find the vertex of the parabola [tex]f(x)=x^2+6x+13[/tex].
[tex]a=1[/tex]
[tex]b=6[/tex]
[tex]h= \frac{-6}{2(1)} [/tex]
[tex]h=-3[/tex]
[tex]k=f(-3)=(-3)^2+6(-3)+13[/tex]
[tex]k=4[/tex]
Now we can use our formula to convert the quadratic function to vertex form:
[tex]f(x)=a(x-h)^2+k[/tex]
[tex]f(x)=1(x-(-3))^2+4[/tex]
[tex]f(x)=(x+3)^2+4[/tex]
We can conclude that the vertex form of the quadratic function is [tex]f(x)=(x+3)^2+4[/tex].
3. Remember that the x-intercepts of a quadratic function are the zeros of the function. To find the zeros of a quadratic function, we just need to set the function equal to zero (replace [tex]f(x)[/tex] with zero) and solve for [tex]x[/tex].
[tex]f(x)=x^2+4x-60[/tex]
[tex]0=x^2+4x-60[/tex]
[tex]x^2+4x-60=0[/tex]
To solve for [tex]x[/tex], we need to factor our quadratic first. To do it, we are going to find two numbers that not only add up to be equal 4 but also multiply to be equal -60; those numbers are -6 and 10.
[tex](x-6)(x+10)=0[/tex]
Now, to find the zeros, we just need to set each factor equal to zero and solve for [tex]x[/tex].
[tex]x-6=0[/tex] and [tex]x+10=0[/tex]
[tex]x=6[/tex] and [tex]x=-10[/tex]
We can conclude that the x-intercepts of the quadratic function [tex]f(x)=x^2+4x-60[/tex] are the points (0,6) and (0,-10).
4. To solve this, we are going to use function transformations and/or a graphic utility.
Function transformations.
- Translations:
We can move the graph of the function up or down by adding a constant [tex]c[/tex] to the y-value. If [tex]c\ \textgreater \ 0[/tex], the graph moves up; if [tex]c\ \textless \ 0[/tex], the graph moves down.
- We can move the graph of the function left or right by adding a constant [tex]c[/tex] to the x-value. If [tex]c\ \textgreater \ 0[/tex], the graph moves left; if [tex]c\ \textless \ 0[/tex], the graph moves right.
- Stretch and compression:
We can stretch or compress in the y-direction by multiplying the function by a constant [tex]c[/tex]. If [tex]c\ \textgreater \ 1[/tex], we compress the graph of the function in the y-direction; if [tex]0\ \textless \ c\ \textless \ 1 [/tex], we stretch the graph of the function in the y-direction.
We can stretch or compress in the x-direction by multiplying [tex]x[/tex] by a constant [tex]c[/tex]. If [tex]c\ \textgreater \ 1[/tex], we compress the graph of the function in the x-direction; if [tex]0\ \textless \ c\ \textless \ 1 [/tex], we stretch the graph of the function in the x-direction.
a. The [tex]c[/tex] value of [tex]f(x)[/tex] is 2; the [tex]c[/tex] value of [tex]g(x)[/tex] is -3. Since [tex]c[/tex] is added to the whole function (y-value), we have an up/down translation. To find the translation we are going to ask ourselves how much should we subtract to 2 to get -3?
[tex]c+2=-3[/tex]
[tex]c=-5[/tex]
Since [tex]c\ \textless \ 0[/tex], we can conclude that the correct answer is: It is translated down 5 units.
b. Using a graphing utility to plot both functions (picture 2), we realize that [tex]g(x)[/tex] is 1 unit to the left of [tex]f(x)[/tex]
We can conclude that the correct answer is: It is translated left 1 unit.
c. Here we have that [tex]g(x)[/tex] is [tex]f(x)[/tex] multiplied by the constant term 2. Remember that We can stretch or compress in the y-direction (vertically) by multiplying the function by a constant [tex]c[/tex].
Since [tex]c\ \textgreater \ 0[/tex], we can conclude that the correct answer is: It is stretched vertically by a factor of 2.
