Asked by Alissa
How do i convert quadratic functions from standard to vertex form?
y=2x^2+12x+14
y=x^2+x+1
y=2x^2+12x+14
y=x^2+x+1
Answers
Answered by
Damon
complete the square
y=2x^2+12x+14
you need the coefficient of x^2 to be 1 so divide everything by 2
y/2 = x^2 + 6 x + 7
move the constant to the left
y/2 - 7 = x^2 + 6 x
add the square of half the coef of x to both sides, in other words 3^2 or 9
y/2 + 2 = x^2 + 6 x + 9
or
y/2 + 2 = (x+3)^2
that is a parabola symmetric about the line x = -3
when x = -3
y/2 + 2 = 0
so y = - 4
so
vertex at (-3 , -4)
y=2x^2+12x+14
you need the coefficient of x^2 to be 1 so divide everything by 2
y/2 = x^2 + 6 x + 7
move the constant to the left
y/2 - 7 = x^2 + 6 x
add the square of half the coef of x to both sides, in other words 3^2 or 9
y/2 + 2 = x^2 + 6 x + 9
or
y/2 + 2 = (x+3)^2
that is a parabola symmetric about the line x = -3
when x = -3
y/2 + 2 = 0
so y = - 4
so
vertex at (-3 , -4)
Answered by
Damon
y = x^2 + 1 x + 1
y-1 = x^2 + 1 x
y-1 + 1/4 = x^2 + x + 1/4
y - 3/4 = (x+1/2)^2
vertex at (-1/2 , 3/4)
y-1 = x^2 + 1 x
y-1 + 1/4 = x^2 + x + 1/4
y - 3/4 = (x+1/2)^2
vertex at (-1/2 , 3/4)
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