Algebra II: Shifting Quadratic Graphs

36-38, shifting quadratic graphs and finding x-intercepts (roots)

Category:

Education

Tags:

• shifting
• quadartic
• graph
• roots

[J2MEZROCK]

A simpler way to factor 12x^2 - 5x - 2 would be to multiply the coefficient of the first term by the third term and then continuing to factor that equation.

x^2 - 5x - 24 = 0

(x-8)(x+3) = 0 Now you bring back the original coefficient and try to factor something out

(12x-8)(12x+3)

4(3x-2)3(4x+1) this is equivalent to just (3x-2)(4x+1) = 0

Ta Da!

[bringintoastyback]

So when you look at the equation, the -2 makes the equation open downwards, and then the vertex is (1,1). It's an easy way to look at it and saves valuable time instead of going through the method of figuring out the original equation then what the shifts are.

[bringintoastyback]

A more simple way to figure out the last question would be to look at the equation itself. y=-2(x-1)^2-1 When it is in this form, you can find the vertex from the equation. Just take the the number inside the bracket and switch the sign to get your x value and take the number outside the bracket on the right to get the y intercept exactly as it is.