Question

Find the Zeros and The vertex of this problem. It’s a graph and I have to show my work for it as well: h(a) = 2a^2 + 3a - 12.

a) Zeros: -4, 3; Vertex: (1.5, -15.25)
b) Zeros: 4, -3; Vertex: (-1.5, -15.25)
c) Zeros: 6, -2; Vertex: (2.5, -13.25)
d) Zeros: -6, 2; Vertex: (-2.5, -13.25)

Answer 1

To find the zeros, set the equation equal to zero and solve. The zeros are -4 and 3. The vertex can be found using the formula x = -b / 2a, which gives a vertex of (1.5, -15.25).

Step-by-step explanation:

To find the zeros of the quadratic equation h(a) = 2a^2 + 3a - 12, we can set h(a) equal to zero and solve for the values of a that make the equation true. We can use factoring or the quadratic formula to solve for the zeros. In this case, the zeros are -4 and 3.

To find the vertex of the quadratic equation, we can use the formula x = -b / 2a, where a and b are the coefficients of the quadratic equation. In this case, the vertex is (1.5, -15.25).