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

Given the equation y = -x^2 + 10x - 27, we can see that a = -1, b = 10, and c = -27.

First, let's find h:
h = -b/2a = -10/(2*-1) = 5

Next, let's find k by substituting h = 5 into the equation:
k = -(5)^2 + 10*5 - 27 = -25 + 50 - 27 = -2

So, the vertex of the parabola is (5, -2).

Question

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

Given the equation y = -x^2 + 10x - 27, we can see that a = -1, b = 10, and c = -27.

First, let's find h:
h = -b/2a = -10/(2*-1) = 5

Next, let's find k by substituting h = 5 into the equation:
k = -(5)^2 + 10*5 - 27 = -25 + 50 - 27 = -2

So, the vertex of the parabola is (5, -2).

Knowee AI · Accepted Answer

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

Given the equation y = -x^2 + 10x - 27, we can see that a = -1, b = 10, and c = -27.

First, let's find h:
h = -b/2a = -10/(2*-1) = 5

Next, let's find k by substituting h = 5 into the equation:
k = -(5)^2 + 10*5 - 27 = -25 + 50 - 27 = -2

So, the vertex of the parabola is (5, -2).

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an left bracket, x, comma, y, right bracket(x,y) point.y, equals, minus, x, squared, plus, 10, x, minus, 27y=−x 2 +10x−27