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5. y = -5(x + 7)^2 - 6 circle one: vertex or standard vertex min or max…

Question

  1. y = -5(x + 7)^2 - 6

circle one: vertex or standard
vertex
min or max
axis of symmetry
y - intercept
direction of opening

Explanation:

Step1: Identify the form

The equation $y = - 5(x + 7)^2-6$ is in vertex - form $y=a(x - h)^2 + k$, where $(h,k)$ is the vertex. Here $h=-7$ and $k = - 6$.

Step2: Determine min/max

Since $a=-5<0$, the parabola opens downwards and has a maximum value. The maximum occurs at the vertex.

Step3: Find axis of symmetry

For a parabola in vertex - form $y=a(x - h)^2 + k$, the axis of symmetry is the vertical line $x = h$. So, the axis of symmetry is $x=-7$.

Step4: Calculate y - intercept

To find the y - intercept, set $x = 0$. Then $y=-5(0 + 7)^2-6=-5\times49-6=-245-6=-251$.

Step5: Determine direction of opening

Since $a=-5<0$, the parabola opens downwards.

Answer:

PropertyValue
Min or MaxMax
Axis of Symmetry$x = - 7$
y - intercept$-251$
Direction of OpeningDownwards