Question

Suppose y partly varies directly as x^(2) and partly varies directly as x. If x=1, then y=14; if x=6, then y=144.
(a) Express y in terms of x.
=> y=2x^(2)+12x
(b) Find the values of x when y=32.
=> x=2/-8
(c) Find the minimum value of y.

i am doubted in c...

Answer 1

pertly varies? what does that mean ....

Answer 2

if a and b are correct ... then the minimum point of a parabola is its vertex

Answer 3

i forgot how to calculate the vertex

Answer 4

y = 2x^2 + 12x

y = 2(x^2 + 6x)

y = 2(x^2 + 6x +9 -9)

y = 2((x+3)^2 -9)

y = 2(x+3)^2 -18

Answer 5

you can either calculate it, or restructure the equation into its vertex form

Answer 6

ohh thanks

Answer 7

the last term, what would normally be a y intercept ... in a quadratic equation is the y value of the vertex

Answer 8

so the minimum is ... ??

Answer 9

-18

Answer 10

-18 is correct

Answer 11

thank you very much:)