Vertex Form

Honelako gehiago

Quadratics Concept Map Assignment

by Rushikesh Amin

Quadratics Assignment

by ALI ZEIDAN

Intervention Map

by Amanda Clapp

Quadratic relations

by Patel Sheil

Vertex Form

Creating an equation in vertex form y = a(x-h)^2 + k

Value k is your optimum value (sign stays the same)

value h is the AoS (when writing it it will be the opposite sign). Ex. If AoS is + h will be negative in the equation

Finding y-intercept

How: Convert to standard form

y = ax^2 + bx + c

Value c is the y-intercept

y = 4(x-1)^2 - 6 expand and simplify y = 4(x-1)(x-1) - 6 y = 4(x^2 - x - x + 1) - 6 y = 4(x^2 - 2x + 1) - 6 y = 4x^2 - 8x + 4 - 6 y = 4x^2 - 8x - 2 y-intercept is (0,-2)

Finding x-intercepts (zeros)

How: Convert to factored form

y = a(x-r)(x-s)

Values r and s are your zeros

When converting from Vertex form to factored form, if the number you are square rooting is negative, then there are no zeros (no factored form).

y = 2/3(x-3)^2 + 5 0 = 2/3(x-3)^2 + 5 -5 = 2/3(x-3)^2 (-5)/(2/3) = (2/3(x-3)^2)/(2/3) -7.5 = (x-3)^2 cannot square root -7.5, therefore there are no zeros

y = 4(x-1)^2 - 6 set y as 0 0 = 4(x-1)^2 - 6 move constant over to other side 6 = 4(x-1)^2 divide each side by value a 6/4 = (4(x-1)^2)/4 1.5 = (x-1)^2 sqrt each side to cancel out ^2 + or - sqrt 1.5 = sqrt (x-1)^2 solve for x +1.22 (rounded) = x-1 x = 2.22 -1.22 = x-1 x = -0.22 x-intercepts are (2.22,0) and (-0.22,0)

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

Example y = 4(x-1)^2 - 6 4 > 0 opens up vertex = (1, -6)

Vertex = values h and k = (h,k)

Special cases

Sometimes the vertex of a parabola will be on the x or y axis. This means that the x or y value of the vertex will be 0. The equation may not have either the value of -h or k

Example

y = 3(x)^2 + 3 vertex = (0,3)

k = optimum value example: y = 4(x-1)^2 - 6 -6 = optimum value

h = AoS (axis of symmetry) example: y = 4(x-1)^2 - 6 x = 1

Direction of opening a > 0 opens up a < 0 opens down