Answer:
[tex]f(x)=9(x+\frac{1}{6})^2-\frac{21}{4}[/tex]
Step-by-step explanation:
So, we need to find the equation of a parabolic function that goes through the points (-3,67) and (-1,1) and has a stretch factor of 9.
In other words, we want to find a quadratic with a vertical stretch of 9 that goes through the points (-3,67) and (-1,1).
To do so, we first need to write some equations. Let's use the vertex form of the quadratic equation. The vertex form is:
[tex]f(x)=a(x-h)^2+k[/tex]
Where a is the leading coefficient and (h,k) is the vertex.
Since a is the leading coefficient, it's also our stretch factor. Thus, let a equal 9.
Also, we have two points. We can interpret them as functions. In other words, (-3,67) means that f(-3) equals 67 and (-1,1) means that f(-1) equals 1. Write the two equations:
[tex]f(x)=a(x-h)^2+k\\f(-3)=67=9((-3)-h)^2+k[/tex]
And:
[tex]f(x)=a(x-h)^2+k\\f(-1)=1=9((-1)-h)^2+k[/tex]
Now, we essentially have a system of equations. Thus, to find the original equation, we just need to solve for the vertex. To do so, first isolate the k term in the second equation:
[tex]1=9(-1-h)^2+k\\k=1-9(-1-h)^2[/tex]
Now, substitute this value to the first equation:
[tex]67=9(-3-h)^2+k\\67=9(-3-h)^2+(1-9(-1-h)^2)[/tex]
And now, we just have to simplify.
First, from each of the square, factor out a negative 1:
[tex]67=9((-1)(h+3))^2+(1-9(((-1)(h+1))^2)[/tex]
Power of a product property:
[tex]67=9((-1)^2(h+3)^2)+(1-9((-1)^2(h+1)^2))[/tex]
The square of -1 is positive 1. Thus, we can ignore them:
[tex]67=9(h+3)^2+(1-9(h+1)^2)[/tex]
Square them. Use the trinomial pattern:
[tex]67=9(h^2+6h+9)+(1-9(h^2+2h+1))[/tex]
Distribute:
[tex]67=(9h^2+54h+81)+(1-9h^2-18h-9)[/tex]
Combine like terms:
[tex]67=(9h^2-9h^2)+(54h-18h)+(81+1-9)[/tex]
The first set cancels. Simplify the second and third:
[tex]67=36h+73[/tex]
Subtract 73 from both sides. The right cancels:
[tex]67-73=36h+73-73\\36h=-6[/tex]
Divide both sides by 36:
[tex](36h)/36=(-6)/36\\h=-1/6[/tex]
Therefore, h is -1/6.
Now, plug this back into the equation we isolated to solve for k:
[tex]k=1-9(-1-h)^2[/tex]
First, remove the negative by simplifying:
[tex]k=1-9((-1)(h+1))^2\\k=1-9((-1)^2(h+1)^2)\\k=1-9(h+1)^2[/tex]
Plug in -1/6 for h:
[tex]k=1-9(-\frac{1}{6}+1)^2[/tex]
Add. Make 1 into 6/6:
[tex]k=1-9(-\frac{1}{6}+\frac{6}{6})^2\\ k=1-9(\frac{5}{6})^2[/tex]
Square:
[tex]k=1-9(\frac{25}{36})[/tex]
Multiply. Note that 36 is 9 times 4:
[tex]k=1-9(\frac{25}{9\cdot4})\\ k=1-\frac{25}{4}[/tex]
Convert 1 into 4/4 and subtract:
[tex]k=\frac{4}{4}-\frac{25}{4}\\ k=-\frac{21}{4}[/tex]
So, the vertex is (-1/6, -21/4).
Now, plug everything back into the very original equation with 9 as a:
[tex]f(x)=a(x-h)^2+k\\f(x)=9(x+\frac{1}{6})^2-\frac{21}{4}[/tex]
And this is our answer :)
