Consider the following quadratic function and its graph. f(x) = x2 – 4x – 21 On a coordinate plane, a parabola opens up. It goes through (negative 4, 8), has a vertex around (2, negative 21), and goes through (8, 8). What is the solution of the following quadratic inequality? x2 – 4x – 21 ≤ 0 A number line going from negative 8 to 8. Open circles are at negative 3 and 7. The line between the circles is shaded. A number line going from negative 8 to 8. Closed circles are at negative 3 and 7. The line between the circles is shaded. A number line going from negative 8 to 8. Open circles are at negative 3 and 7. The line to the left of negative 3 and to the right of 7 is shaded. A number line going from negative 8 to 8. Closed circles are at negative 3 and 7. The line to the left of negative 3 and to the right of 7 is shaded.

THE ANSWER IS B

Step-by-step explanation:

ON EDGE

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3