Question

how does the equation h(x)=-(x,-3)2 + 4 compare to the graph of g(x)=x2?

Answer 1

Answer:

option A

Step-by-step explanation:

The curve h(x)=-(x,-3)2 + 4 has a negative coeff for the degree e term. So the curve ud opened downward

To express the function in normal form,

H = -SQ(X) + 6X - 5

So the y intercept is -5

To find the roots of H

H = -SQ(X) + 6X - 5

(x-1)(x-5) = 0

so the curve has x intercept at 1 and 5

Thus the max point is at x=3

so the curve is shift right by 3 units

{ the function g(x)=x2 has turning point (0, 0) }

since the max point is at x=3, sub x=3 into the function H,

the max point is (3, 4)

So the function H is shift up by 4 unit