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The graph of the function f(x) = ax^2 + bx + c has vertex at (0, 2) and passes through the point
(1, 8). Find a, b and c

Respuesta :

Answer:

Step-by-step explanation:

You need to use vertex form of a quadratic to solve this.

Consider the vertex to be [tex](h,k)[/tex]

Another way of representing a quadratic is in "vertex form":

[tex]f(x) = a(x-h)^2+k[/tex]

Now all you have to do is solve for a.  You know that the vertex is [tex](0,2)[/tex] and you have know the point of [tex](1,8)[/tex].  Now, all you have to do is plug in these values and solve for a.

[tex]8 = a(1-0)^2+2\\8=a(1)^2+2\\8=a+2\\a=6[/tex]

Now you know the equation is [tex]f(x) = 6(x-0)^2+2[/tex] , but you need it in quadratic form.  All you have to do is solve is distribute the 6:

[tex]6x^2+2[/tex]

You get:

a = 6

b = 0

c = 2

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