Question

Part 1: What mistake did AJ make in the graph?

Answer 1

Answer:

Part 1) AJ drawn the parabola opening upwards, instead of drawing it opening downwards

Part 2) see the explanation

Step-by-step explanation:

Part 1) What mistake did AJ make in the graph?

we have

$f(x)=-(x+2)^2-1$

This is the equation of a vertical parabola written in vertex form

The parabola open downward (because the leading coefficient is negative)

The vertex represent a maximum

The vertex is the point (-2,-1)

therefore

AJ drawn the parabola opening upwards, instead of drawing it opening downwards

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?

take the values x=-4 and x=4

For x=-4

substitute the value of x in the quadratic equation

$f(x)=-(-4+2)^2-1\\f(x)=-5$

For x=4

substitute the value of x in the quadratic equation

$f(x)=-(4+2)^2-1\\f(x)=-37$

According to AJ's graph for the value of x=-4 the function should be positive, however it is negative and for the value of x=4 the function should be positive and the function is negative

therefore

AJ made a mistake in the graph