Which of the following equations has a maximum at (9,7)
a. y = -12 + 141 - 40
b. y = -12 -180 -88
c. y = -12 + 18a - 74
d. y = –12 – 14. – 58

Step-by-step explanation:

option C ⇒ C) y = -x² + 18x - 74

Step-by-step explanation:

The given options are:

A) y = -x² + 14x - 40

B) y = -x² - 18x - 88

C) y = -x² + 18x - 74

D) y = -x² - 14x - 58

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The general equation of the parabola has the form:

y = f(x) = ax² + bx + c

The vertex of the parabola has the coordinates (h , k)

where h =

and     k = f(h) = f()

Check option A: a = -1 , b = 14  ⇒ (h,k) = (7, f(7) ) = (7 , 9)

Check option B: a = -1 , b = -18 ⇒ (h,k) = (-9, f(-9) ) = (-9,-7)

Check option C: a = -1 , b = 18  ⇒ (h,k) = (9, f(9) ) = (9,7)

Check option D: a = -1 , b = -14 ⇒ (h,k) = (-7, f(7) ) = (-7,-9)

So the equation which has a maximum at (9,7) ⇒ y = -x² + 18x - 74

So, the correct answer is option C

C

Step-by-step explanation:

y = a(x - h)² + k

Since it's a maximum turning point:

a < 0

From the options available, a = -1

y = -(x - 9)² + 7

y = -(x² - 18x + 81) + 7

y = -x² + 18x - 81 + 7

y = -x² + 18x - 74

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.

look at the picture.

y-intercept for the function on a picture is 4 → point (0, 4).

y-intercept for the function

is for x = 0. substitute:

y-intercept is 8 → point (0, 8).