Find the quadratic equation whose parabola has vertex (3,-2) and y-intercept (0, 16). Give your answer in vertex form.

y = 2*(x - 3)^2 - 2

Step-by-step explanation:

Remember that a quadratic equation of vertex (h, k) is written as:

y = a*(x - h)^2 + k

Where a is the leading coefficient.

So, if we know that the vertex is at (3, - 2)

we have:

y = a*(x - 3)^2 + (-2)

And we want the y-intercept to be (0, 16)

This means that, when we take x = 0, we must have y = 16

if we replace these in the above equation we get:

16 = a*(0 - 3)^2 - 2

now we can solve this for a

16 = a*(-3)^2 - 2

16 = a*9 - 2

16 + 2 = a*9

18 = a*9

18/9 = a

2 = a

Then the quadratic equation is:

y = 2*(x - 3)^2 - 2

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

{3 + i, 3 - i, 3, -1}

step-by-step explanation: