Question

A.) Determine "a" and "k" so both points are on the graph of the function.

1.) (0,1) (2,1); y=a(x-1)^2+k
2.) (1,11) (2,-19); y=a(x+1)^2+k


B.) Determine whether or not the function f(x)=0.25(2x-15)^2+15 has a mazimum or a minimum value. Then find the value.


C.) When y=2(x-3)(x+5) is written in standard form, what is the value of b?


D.) For y=3x^2-7x+5, what is the x-value of the vertex? Enter your answer as an improper fraction in simplest form.


THANK YOU SO MUCH!!!!! :)

Answer 1

A1
1 = a(-1)^2 + k
1 = a(1)^2 + k
----------------
-1 = 0 + 0 no solution possible

A2
11 = a(2^2) + k
-19 = a(3^2) + k
-----------------subtract
30 = -5 a
a = -6
11 = -6(4) + k
k = 35
so
y = -6(x+1)^2 + 35

Answer 2

B.) Determine whether or not the function f(x)=0.25(2x-15)^2+15 has a mazimum or a minimum value. Then find the value.

when x gets big + or big -, y gets big +
SO
This parabola has a minimum (holds water)
it is symmetric about the point where
2x-15 = 0
or x = 7.5
then calculate y at x = 7.5 to get the vertex

Answer 3

C.) When y=2(x-3)(x+5) is written in standard form, what is the value of b?

y = 2(x^2 + 2 x - 15)

y = 2 x^2 + 4 x - 30

Answer 4

D.) For y=3x^2-7x+5, what is the x-value of the vertex? Enter your answer as an improper fraction in simplest form.

3 x^2 - 7 x = y - 5

x^2 -(7/3) x = y/3 - 5/3

x^2-(7/3)x+(7/6)^2 = y/3 -5/3+(7/6)^2

(x-7/6)^2 = y/3 etc etc
so
when x = 7/6 we are at the vertex