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

1 Answer
May 14, 2018

The vertex is (-2/5,41/5) or (-0.4,8.2).

Explanation:

Given:

y=4x^2-2x+(x+3)^2

First you need to get the equation into standard form.

Expand (x+3)^2 using the FOIL method.
https://www.mathsisfun.com/definitions/foil-method.html

y=4x^2-2x+x^2+6x+9

Collect like terms.

y=(4x^2+x^2)+(-2x+6x)+9

Combine like terms.

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

ax^2+bx+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=(-b)/(2a)

x=(-4)/(2*5)

x=-4/10

Simplify.

x=-2/5 or -0.4

To calculate the y-coordinate of the vertex, substitute -2/5 for x in the equation and solve for y.

y=5(-2/5)^2+4(-2/5)+9

y=5(4/25)-8/5+9

y=20/25-8/5+9

Simplify 20/25 to 4/5.

y=4/5-8/5+9

Multiply 9 by 5/5 to get an equivalent fraction with 5 as the denominator.

y=4/5-8/5+9xx5/5

y=4/5-8/5+45/5

Simplify.

y=41/5 or 8.2

The vertex is (-2/5,41/5) or (-0.4,8.2).

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