Given the function g(x)=square root(-2x+6) and h(x)=2x^2-3x+a, evaluate h(g(2))

Question

aum · Answer

$$
g(x) = \sqrt{-2x+6} \
g(2) = \sqrt{-2*2+6} = ?\
h(x) = 2x^2 - 3x + a 
$$Take the output of g(2), plug it into h( g(2) ) and evaluate.

anonymous · Answer

so i found a=.28, so then do i plug .28 into h(x) and solve? when i did that i got h(g(2))=0 @aum

aum · Answer

How did you find 'a' ?

anonymous · Answer

i found that g(2)=1.4 and plugged that in for h(x) and found 'a'. then i plugged in the value of a and g(2) and got 0

aum · Answer

$$
g(x) = \sqrt{-2x+6} \
g(2) = \sqrt{-2*2+6} = \sqrt{-4+6} = \sqrt{2}\
h(x) = 2x^2 - 3x + a \
h(g(2)) = h(\sqrt{2}) = 2*(\sqrt{2})^2 - 3*\sqrt{2} + a = 2*2 - 3\sqrt{2} + a \
h(g(2)) = 4 - 3\sqrt{2} + a
$$

anonymous · Answer

oh, tht makes sence, i did too much work