What is the the vertex of y = (x-3)(x-4) +4+12x ?

2 Answers
Apr 4, 2018

The coordinates of the vertex are (-5/2, 39/4).

Explanation:

y=(x-3)(x-4)+4+12x

Let's put this in standard form first. Expand the first term on the right-hand side using the distributive property (or FOIL if you like).

y=x^2-7x+12+4+12x

Now combine like terms.

y=x^2+5x+16

Now complete the square by adding and subtracting (5/2)^2 to the right-hand side.

y=x^2+5x+25/4+16-25/4

Now factor the first three terms of the right-hand side.

y=(x+5/2)^2+16-25/4

Now combine the last two terms.

y=(x+5/2)^2+39/4

The equation is now in vertex form

y=a(x-k)^2+h

In this form, the coordinates of the vertex are (k, h).

Here, k=-5/2 and h=39/4, so the coordinates of the vertex are (-5/2, 39/4).

Apr 5, 2018

The vertex is (-5/2,39/4) or (-2.5,9.75).

Explanation:

Given:

y=(x-3)(x-4)+4+12x

First get the equation into standard form.

FOIL (x-3)(x-4).
https://www.ipracticemath.com/learn/algebra/foil-method-of-binomial-multiplication

y=x^2-7x+12+4+12x

Collect like terms.

y=x^2+(-7x+12x)+(12+4)

Combine like terms.

color(blue)(y=x^2+5x+16 is a quadratic equation in standard form:

y=ax^2+bx+c,

where:

a=1, b=5, c=16

The vertex is the maximum or minimum point of a parabola. The x coordinate can be determined by using the formula:

x=(-b)/(2a)

x=(-5)/(2*1)

x=-5/2=-2.5

To find the y coordinate, substitute -5/2 for x and solve for y.

y=(-5/2)^2+5(-5/2)+16

y=25/4-25/2+16

Multiply 25/2 and 16 by fractional forms of 1 to convert them to equivalent fractions with the denominator 4.

y=25/4-25/2xx2/2+16xx4/4

y=25/4-50/4+64/4

y=(25-50+64)/4

y=39/4=9.75

The vertex is (-5/2,39/4) or (-2.5,9.75).

graph{y=x^2+5x+16 [-13.5, 11.81, 6.47, 19.12]}