Question

find the values of a h and k that make the equation 1/2x^2-3x+5=a(x-h)+k

Answer 1

ok here we go. best way to do this is compute
\[\frac{-b}{2a}\]

Answer 2

it is going to be
\[\frac{1}{2}(x-h)^2+k\]

Answer 3

im confused is it that simple?

Answer 4

\[h=\frac{b}{2a}=\frac{-3}{2\times \frac{1}{2}}=-3\]

Answer 5

so you will have
\[\frac{1}{2}(x-3)^2+k\]

Answer 6

mmhmm

Answer 7

now to find k, you see that if you replace x by 3 is (x-3) you will get 0+k=k so replace x by 3 in the original equation to find k

Answer 8

yes it is that simple. usually made to seem harder than it is

Answer 9

\[h=\frac{b}{2a}\] and to find k replace x by
\[-\frac{b}{2a}\] in the original expression

Answer 10

this is how you find the vertex

Answer 11

you have
\[y=\frac{1}{2}x^2-3x+5\]
\[y=\frac{1}{2}3^3-3\times 3+5\]
\[y=\frac{9}{2}-9+5\]
\[y=-\frac{9}{2}+5\]
\[y=\frac{1}{2}\]

Answer 12

so your answer is
\[\frac{1}{2}(x-3)^2+\frac{1}{2}\]

Answer 13

this is a lot easier than completing the square and trying to figure out what you added and subtracted. it is a straightforward computation

Answer 14

easy yes? i mean relatively easy