Question

In quadratic equations, how do you convert an equation from general form (y = a(x-h)^2+k) to standard form (y = ax^2+bx+c)?

Answer 1

Please show work!

Answer 2

Square the binomial. Then distribute the 1/2. Then combine like terms and write it in descending order of degree.

Answer 3

Where does the x come from in the second monomial?
1/2(x-3)^2+8

Answer 4

So far I have y = 1/2x^2 - 4.5 + 8? Or is there no x?

Answer 5

\(y = \dfrac{1}{2}(x - 3)^2 + 8\)

First, square the binomial x - 3.
The square of a binomial has 3 terms and follows the pattern:
\( (a + b)^2 = a^2 + 2ab + b^2\)
\( (a - b)^2 = a^2 - 2ab + b^2\)
You need the second one because you have x - 3, not x + 3.

Answer 6

In general,
\((a + b)^2 \ne a^2 + b^2 \)
\((a - b)^2 \ne a^2 - b^2 \)

Answer 7

So that would be 1/2(x^2 - 6x - 9) + 8
Which would be 1/2x^2 - 3x - 4.5 +8
Then you would get 1/2x^2 - 3x + 3.5
Right?

Answer 8

Almost. The sign of the 9 is +9, not -9 when you square the binomial.

Answer 9

Okay, thanks!

Answer 10

I understand how to do it now.

Answer 11

\(y = \dfrac{1}{2}(x - 3)^2 + 8\)

\(y = \dfrac{1}{2}(x^2 - 6x + 9) + 8 \)

\(y = 0.5x^2 - 3x + 4.5 + 8 \)

\(y = 0.5x^2 - 3x + 12.5 \)

Answer 12

Great.