Question

AJ graphs the function LaTeX: f\left(x\right)=-\left(x+2\right)^2-1f ( x ) = − ( x + 2 ) 2 − 1 as shown below. On a coordinate plane, a parabola opens up. It goes through (negative 3, 0), has a vertex at (negative 2, negative 1), and goes through (negative 1, 0). A vertical dashed line at (negative 2, 0) is parallel to the y-axis. Part 1: What mistake did AJ make in the graph? Part 2: Evaluate any two x-values (between -5 and 5) into AJ's function. Show your work. How does your work prove that AJ made a mistake in the graph?

Answer 1

x = - 1 and - 3

Explanation:

Answer 2

Part 1 : The parabola opens down and it does not passes through the points (-3,0) and (-1,0) points.

Part 2 : (-3,-2) and (-1,-2) are the two values on the parabola.

Explanation:

The general equation of a parabola with vertex at (α, β) and axis parallel to negative y-axis is given by

(x - α)² = - 4a(y - β) ............ (1)

Part 1 : Now, the given parabola is f(x) = y = - (x + 2)² - 1

⇒ (x + 2)² = - (y + 1) ......... (2)

Now, comparing equations (1) and (2), the vertex of the parabola is (-2,-1) and it opens down. and the axis of the parabola is the line x = - 2.

Now, it does not pass through the points (-3,0) and (-1,0) points.

Part 2 : Now, (-3,-2) and (-1,-2) are the two values on the parabola (2).

Because, it we put y = - 2, in equation (2) then we will get

(x + 2)² = - (- 2 + 1) = 1

⇒ x + 2 = ± 1

⇒ x = - 1 and - 3. (Answer)