Question

Determine the quadratic function f whose graph is given. The vertex is (1,-3) and y-intercept is -2

Answer 1

Answer:

Vertex form: $f(x)=(x-1)^2-3$

Standard form: $f(x)=x^2-2x-2$

Step-by-step explanation:

A quadratic function in vertex form is $f(x)=a(x-h)^2+k$ where $(h,k)$ is the vertex.

We are given $h=1,k=-3$.

Let's plug that in:

$f(x)=a(x-1)^2-3$.

Now let's find $a$.

We will use the $y$-intercept $(0,-2)$ to find $a$.

$f(0)=a(0-1)^2-3$

$-2=a(0-1)^2-3$

$-2=a(-1)^2-3$

$-2=a(1)-3$

$-2=a-3$

$1=a$

So the function in vertex form is:

$f(x)=(x-1)^2-3$.

In standard form, we will have to multiply and combine any like terms.

Let's do that:

$f(x)=(x-1)(x-1)-3$

$f(x)=x^2-x-x+1-3$

$f(x)=x^2-2x-2$