Question

Find the vertex of y=4x^2+4x-15

Answer 1

`\left(-(1)/(2),\:-16\right)`

Explanation:

The general equation of a parabola in polynomial form is
y = ax² + bx + c

where a, b, c are constants

If a is > 0, the parabola opens upward and the vertex is a minimum
If a < 0, the the parabola opens downward and the vertex is a maximum

The vertex x value is given by the points

`x_v = -b/2a\\`

The given parabola equation is
y = 4x² + 4x - 15

Since we have a = 4, b = 4

`x_v = -b/2a = -{(4)/(2\cdot 4) = - (4)/(8) = - (1)/(2)`

To find the y value of the vertex,
`y_v` , plug this value into the parabola equation and solve for y

`y_v=4\left(-(1)/(2)\right)^2+4\left(-(1)/(2)\right)-15\\\\= 4 \cdot (1)/(4) -2 - 15\\\\= 1 - 2 - 15\\\\= -16`

So the vertex is at
`\left(-(1)/(2),\:-16\right)` and is a minimum