Question

Rewrit the equation in vertex form. Name the vertex and y-intercept.
y=3/5^2+30x+382
I need some help do not understand these type of problems.

Answer 1

y = 3/5 x^2 + 30x + 382
= 3/5 (x^2 + 50x) + 382
Now come s the tricky bit. Take 1/2 of 50 and square it. Add it inside the parens, and subtract it outside the parens. Be sure to recall the 3/5 multiplier involved.
y = 3/5 (x^2 + 50x + 625) - 3/5(625) + 382
y = 3/5 (x+25)^2 + 7
(y-7) = 3/5 (x+25)

And behold: standard vertex form for a parabola.

vertex at (-25,7)
y-intercept at 382 (but we knew that from the original equation, when x=0)

Answer 2

oops. dropped an exponent:
(y-7) = 3/5 (x+25)^2

Answer 3

Thank you for explaining problems so that one can understand. Very much appreciated.

Answer 4

He is 100% correct.