Question

Graph he function. F(x)=-x^2+2x+4

Answer 1

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

Answer 2

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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