Question

# 9. Consider the graph y=x²+12x+32 a. Write the equation of the axis of symmetry. b. Find the coordinates of the vertex (0₁-20) c. Identify the vertex as a minimum maximum d. Graph the function​

Answer 1

a) x = -6

b) (-6, -4)

c) Minimum

d) See the graph below.

Explanation:

a)

We are given a parabola as y = x² + 12x + 32. The axis of symmetry from a parabola is given as x = -b/2a when y = ax² + bx + c. This means our axis of symmetry will be:

x = -b/2a x = -12/2(1) x = -6

b)

The vertex is the lowest or highest point of the parabola. This given function is not in vertex form, but we can still find the vertex without changing the form.

The vertex is the point of intersection of the parabola and the axis of symmetry, so we can substitute -6 into the function to solve for y.

y = x² + 12x + 32 y = (-6)² + 12(-6) + 32 y = -4

This means our vertex is (-6, -4).

c)

The vertex will be the minimum if the parabola opens down and the maximum if the parabola opens up. Since the coefficient of x² is positive, this parabola will open up. This makes our vertex the minimum.

d)

See the graph below. Start by plotting the vertex, then graph the parabola opening up around x = -6.