Use tl con
15 When the function g(x) =
5x, x < 3
x2 + 4, x > 3
is graphed correctly, how
should the points be drawn on the graph for an x-value of 3?
(1) open circles at (3,15) and (3,13)
(2) closed circles at (3,15) and (3,13)
(3) an open circle at (3,15) and a closed circle at (3,13)
(4) a closed circle at (3,15) and an open circle at (3,13)

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

step-by-step explanation:

answer: c

substitute in the values.

for a:

f(0) = 0 + 9

f(0) = 9

a is a solution.

for b:

f(7) = 14 + 9

f(7) = 23

b is a solution.

for c:

f (3) = 6 + 9

f(3) = 15

c is not a solution. it's supposed to be 14.

for d:

f(2) = 4 + 9

f(2)= 13

d is a solution.