Question

I need help with this question please. Ignore the wording below, it’s also from a homework practice

Answer 1

The form of the quadratic function is

`f(x)=ax^2+bx+c`

Where c is the y-intercept, value y at x = 0

From the given picture

There are 3 points lie on f(x) (0, 3), (1, 2), (3, -12)

Since at x = 0, y = 3, then

The y-intercept is (0, 3), then

c = 3

The function could be A, or C, or D

Then we will use the point (1, 2) to find which of them is the answer

Substitute x by 1 in equation A to find f(x) (value of y)

`\begin{gathered} x=1 \\ f(1)=2(1)^2+1+3 \\ f(1)=2+1+3 \\ f(1)=6 \\ y=6,atx=1 \end{gathered}`

But at x = 1, y = 2

Then A is wrong

Substitute x by 1 in equation C

`\begin{gathered} x=1 \\ f(1)=-2(1)^2+1+3 \\ f(-1)=-2+1+3 \\ f(-1)=2 \\ y=2 \end{gathered}`

Since the value of y equal the y-coordinate at x = 1

Then C is correct

The function that contains points (0, 3), (1, 2), and (3, -12) is C