Question

These 4 on my Review... STUCK! Help?
equation using ------> f(x)= ax^2 + bx+c

1) Write the equation of the quadratic function with roots -1 and 1 and a vertex at (0, -3).
2) Write the equation of the quadratic function with roots -10 and -8 and a vertex at (-9, -3).
3) Write the equation of the quadratic function with roots 0 and 2 and a vertex at (1, 5).
4) Write the equation of the quadratic function with roots 6 and 10 and a vertex at (8, 2)

Answer 1

well first of all each of these has tmi, but no matter, lets do one and all the rest are identical
now you have a choice.
you can use the vertex form first, or the factored form first. which one do you pick?

Answer 2

how about factored?.... whatever is the best/simplest way.

Answer 3

ok we can do one of each
factored means if the zeros are say \(r_1\) and \(r_2\) then you start by writing
\[f(x)=a(x-r_1)(x-r_2)\] and then solve for \(a\) second, using the other point
lets do the second one that way

Answer 4

roots are -10 and -8 so you start by writing
\[f(x)=(x-(-10))(x-(-8))=a(x+10)(x+8)\]

Answer 5

now you know \((-9,-3)\) is on the graph, so you can replace \(x\) by \(-9\) and \(f(x)\) by \(-3\) and solve for \(a\)

Answer 6

you get
\[f(-9)=a(-9+10)(-9+8)=-3\]
\[-a=-3\]\[a=3\]

Answer 7

so your answer is
\[f(x)=3(x+10)(x+8)\] and you can multiply out if you like

Answer 8

I only need an equation for them, but I kind of get it.... still a bit lost :/

Answer 9

you will notice that i did not use the fact that the point \((-9,-3)\) was the vertex, just that it is another point on the graph, we can do it a second way if you like

Answer 10

bleh. let me look at this one..... im stumped still :/

Answer 11

ok i skipped no steps, it is all there