4 years ago

Find the range of the function f(x)=10-x squared

1 answer:

4 0

F’(x) = d over dx (10-xsq

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Answer:

D) $(-2,-9)$

Step-by-step explanation:

Vertex Form of a Vertical Parabola:

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

Vertex -> $(h,k)$

Axis of Symmetry -> $x=h$

Vertical Scale Factor -> $a$

To turn $f(x)=x^2+4x-5$ into vertex form, we need to complete the square on the right side
Therefore, if $f(x)=0$ , then $0+9=x^2+4x-5+9$ completes the square on the right side
This becomes $9=(x+2)^2$
This means that our function in vertex form is $f(x)=(x+2)^2-9$