Vertex Form

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Quadratic Equations By. Eshal Khaliq

Khaliq Eshalにより

Introduction to functions.

Mya Morgan - Sandalwood Heights SS (2442)により

Algebra A

Daniel Slavovにより

Quadratic relations

Patel Sheilにより

Find x values

y = 4(x-1)^2 - 6

0 = 4(x-1)^2 - 6 6 = 4(x-1)^2 6/4 = (4(x-1)^2)/4 1.5 = (x-1)^2 sqrt + or - 1.5 = sqrt (x-1)^2 1.22 (rounded) = (x-1) 1.22 + 1 =x x = 2.22 -1.22 = (x-1) -1.22 + 1 = x x = -0.22

x ints are = 2.22 and -0.22 y = 4(x-2.22)(x+0.22)

set y as 0 and solve for x

Remember to use samdeb and not bedmas when factoring

When writing the Factored form with the x-int, r and s are equal to x-ints but the opposite signs (if the x-int is positive,it will be -r or -s, if x-int is negative, it will be +r or +s)

Take value c to other side

Divide both sides by value a

Square root both sides

Don't forget the positive and negative roots

Solve for x

Expand and simplify

y = 2/3(x-3)^2 + 5

Expand (x-3)^2 y = 2/3(x-3)(x-3) + 5 Multiply (x-3)(x-3) using FOIL y = 2/3(x^2 - 3x - 3x + 9) + 5 combine like terms in brackets y = 2/3x^2 - 4x + 6 + 5 combine like terms y = 2/3x^2 - 4x + 11

Don't forget to use Use Foil (multiply firsts, outsides, insides and lasts

Vertex Form

Standard Form

y = ax^2 + bx + c

Factored Form

Special occasions

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). The vertex is either above the x-axis and opening upwards, or it is below and opening downwards.

Example

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

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

A value is the same in each form