Question

f(x)=-x^2+10+5

Answer 1

the vertex of parabola i got was (-5,-20)
f(x)=-(x+5)^2-20

Answer 2

the x-intercept i got was (-0.5,0)(-10.5,0)
because i changed f(x) into 0
and got 0=-x^2+10x+5
plugged it in the quadratic formula and got the results

Answer 3

but when i tried to solve for the y-intercept
i let x=0
f(0)=(0)^2+10(0)+5
and got (0,5) it looks weird when i graph it like that because the y-axis should touch point (0,-45) because a is -1 and if a is negative the graph should be narrower not wider

Answer 4

f(x)=-a(x-h)^2+k is a narrower graph
f(x)=a(x-h)^2+k is a wide graph

Answer 5

@RadEn

Answer 6

have to go to class, any advises will help

Answer 7

f(x)=-x^2+10x+5
first, take negative sign infront all terms, it can be
f(x) = - (x^2 -10x - 5)

Answer 8

by using complete square we have :
f(x) = - (x^2 -10x - 5)
f(x) = - (x^2 -10x + (1/2 * -10)^2 - (1/2 * -10)^2 - 5)
f(x) = - (x^2 -10x + 25 - 25 - 5)
f(x) = - (x^2 -10x + 25 - 30)
f(x) = - (x^2 -10x + 25) + 30
f(x) = - (x - 5)^2 + 30

that should be easier for you now