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

Given the equation of the parabola y = 5x^2 - 30x + 45, the coefficients are a = 5, b = -30, and c = 45.

Step 1: Calculate h
h = -b/2a = -(-30)/(2*5) = 30/10 = 3

Step 2: Substitute h into the equation to find k
k = 5*(3)^2 - 30*3 + 45 = 45 - 90 + 45 = 0

So, the vertex of the parabola is at the point (3, 0).

Question

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

Given the equation of the parabola y = 5x^2 - 30x + 45, the coefficients are a = 5, b = -30, and c = 45.

Step 1: Calculate h
h = -b/2a = -(-30)/(2*5) = 30/10 = 3

Step 2: Substitute h into the equation to find k
k = 5*(3)^2 - 30*3 + 45 = 45 - 90 + 45 = 0

So, the vertex of the parabola is at the point (3, 0).

Knowee AI · Accepted Answer

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

Given the equation of the parabola y = 5x^2 - 30x + 45, the coefficients are a = 5, b = -30, and c = 45.

Step 1: Calculate h
h = -b/2a = -(-30)/(2*5) = 30/10 = 3

Step 2: Substitute h into the equation to find k
k = 5*(3)^2 - 30*3 + 45 = 45 - 90 + 45 = 0

So, the vertex of the parabola is at the point (3, 0).

Consider the equation of parabola y=5x2−30x+45y=5x 2 −30x+45.Its vertex is located at (( ,, ))