Graph he function. F(x)=-x^2+2x+4
2 answers:
we are given
x-intercept:
we can set f(x)=0
and then we can solve for x
we get
y-intercept:
we can set x=0 and find y
vertex:
now, we can find y-value
so, we get vertex as (1,5)
now, we can draw graph
we are given
f(x)=-x^2+2x+4
x-intercept:
we can set f(x)=0
and then we can solve for x
-x^2+2x+4=0
we get
x=-\sqrt{5}+1,\:x=1+\sqrt{5}
y-intercept:
we can set x=0 and find y
y=-(0)^2+2*0+4
y=4
vertex:
x=-\frac{b}{2a}
x=-\frac{2}{2*-1}
x=1
now, we can find y-value
y=-(1)^2+2*1+4
y=5
so, we get vertex as (1,5)
now, we can draw graph
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