What is the vertex of  y=4x^2-2x+(x+3)^2 ?

May 14, 2018

The vertex is $\left(- \frac{2}{5} , \frac{41}{5}\right)$ or $\left(- 0.4 , 8.2\right)$.

Explanation:

Given:

$y = 4 {x}^{2} - 2 x + {\left(x + 3\right)}^{2}$

First you need to get the equation into standard form.

Expand ${\left(x + 3\right)}^{2}$ using the FOIL method.
https://www.mathsisfun.com/definitions/foil-method.html

$y = 4 {x}^{2} - 2 x + {x}^{2} + 6 x + 9$

Collect like terms.

$y = \left(4 {x}^{2} + {x}^{2}\right) + \left(- 2 x + 6 x\right) + 9$

Combine like terms.

$y = 5 {x}^{2} + 4 x + 9$ is a quadratic equation in standard form:

$a {x}^{2} + b x + c$,

where:

$a = 5$, $b = 4$, $c = 9$

The vertex is the maximum or minimum point of a parabola. Since $a > 0$, the vertex is the minimum point of this parabola, and the parabola opens upward.

The x-coordinate of the vertex is the same as the axis of symmetry for a quadratic equation in standard form. The formula is:

$x = \frac{- b}{2 a}$

$x = \frac{- 4}{2 \cdot 5}$

$x = - \frac{4}{10}$

Simplify.

$x = - \frac{2}{5}$ or $- 0.4$

To calculate the y-coordinate of the vertex, substitute $- \frac{2}{5}$ for $x$ in the equation and solve for $y$.

$y = 5 {\left(- \frac{2}{5}\right)}^{2} + 4 \left(- \frac{2}{5}\right) + 9$

$y = 5 \left(\frac{4}{25}\right) - \frac{8}{5} + 9$

$y = \frac{20}{25} - \frac{8}{5} + 9$

Simplify $\frac{20}{25}$ to $\frac{4}{5}$.

$y = \frac{4}{5} - \frac{8}{5} + 9$

Multiply $9$ by $\frac{5}{5}$ to get an equivalent fraction with $5$ as the denominator.

$y = \frac{4}{5} - \frac{8}{5} + 9 \times \frac{5}{5}$

$y = \frac{4}{5} - \frac{8}{5} + \frac{45}{5}$

Simplify.

$y = \frac{41}{5}$ or $8.2$

The vertex is $\left(- \frac{2}{5} , \frac{41}{5}\right)$ or $\left(- 0.4 , 8.2\right)$.

graph{y=5x^2+4x+9 [-11.72, 13.59, 5.72, 18.38]}