The vertex of a parabola given by the equation y = ax^2 + bx + c is given by the point (h, k), where h = -b/2a and k = c - b^2/4a.

In this case, a = -3, b = -6, and c = 5.

First, let's find h:
h = -b/2a = -(-6)/2*(-3) = 6/-6 = -1

Next, let's find k:
k = c - b^2/4a = 5 - (-6)^2/4*(-3) = 5 - 36/-12 = 5 - (-3) = 5 + 3 = 8

So, the vertex of the parabola y = -3x^2 - 6x + 5 is (-1, 8).

Question

The vertex of a parabola given by the equation y = ax^2 + bx + c is given by the point (h, k), where h = -b/2a and k = c - b^2/4a.

In this case, a = -3, b = -6, and c = 5.

First, let's find h:
h = -b/2a = -(-6)/2*(-3) = 6/-6 = -1

Next, let's find k:
k = c - b^2/4a = 5 - (-6)^2/4*(-3) = 5 - 36/-12 = 5 - (-3) = 5 + 3 = 8

So, the vertex of the parabola y = -3x^2 - 6x + 5 is (-1, 8).

Knowee AI · Accepted Answer

The vertex of a parabola given by the equation y = ax^2 + bx + c is given by the point (h, k), where h = -b/2a and k = c - b^2/4a.

In this case, a = -3, b = -6, and c = 5.

First, let's find h:
h = -b/2a = -(-6)/2*(-3) = 6/-6 = -1

Next, let's find k:
k = c - b^2/4a = 5 - (-6)^2/4*(-3) = 5 - 36/-12 = 5 - (-3) = 5 + 3 = 8

So, the vertex of the parabola y = -3x^2 - 6x + 5 is (-1, 8).

Findthevertexoftheparabolay = –3x2 − 6x + 5.Simplify both coordinates and write them as proper fractions, improper fractions, or integers.