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

Mathematics

1 answer:

Ann [662]2 years ago

7 0

Answer:

Step-by-step explanation:

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

Consider the vertex to be $(h,k)$

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

$f(x) = a(x-h)^2+k$

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

$8 = a(1-0)^2+2\\8=a(1)^2+2\\8=a+2\\a=6$

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

$6x^2+2$

You get:

a = 6

b = 0

c = 2

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