Question

The graph of the function f(x)=ax^2+bx+c has its vertex at (0,1) and passes through the point (1,9) find a b and c

Answer 1

Answer:

$a=8$

$b=0$

$c=1$

Step-by-step explanation:

The vertex form of a quadratic is $y=a(x-h)^2+k$ where the vertex is $(h,k)$.

We are given $h=0,k=1$.

Plugging this in gives us:

$y=a(x-0)^2+1$

Simplifying:

$y=ax^2+1$

Since it passes through $(x,y)=(1,9)$, we can use this information to find $a$.

$9=a(1)^2+1$

$9=a+1$

$8=a$

So the quadratic in vertex form is $y=8x^2+1$.

That happens to be in standard form as well.

$a=8$

$b=0$

$c=1$