What is the standard form of the equation y = 1/2 (x - 4)^2 + 6A) y= 1/2 x^2 − 8x + 22 B) y= 1/2 x^2 − 4x + 22C) y= 1/2 x^2 − 4x + 14D) y= 1/2 x^2 − 8x + 14Which equation represents the quadratic parent function shifted to the left two units? A) y=(x−2)^2 B) y=(x+2)^2 C) y=x^2 − 2 D) y=x^2 + 2Find the equation, in vertex form, for a parabola with a vertex of (2,-5) that goes through the point (3,4).y=.36(x+2)^2−5y=2(x−2)^2−5y=.75(x−2)^2−5 y=9(x−2)^2−5
Accepted Solution
A:
1) The standard form for any equation is [tex]y = a{x}^{2} + bx + c[/tex] with this said we have to get the equation that they gave us into this form so we have [tex]y = \frac{1}{2} {(x - 4)}^{2} + 6[/tex] this means that the order of operations we have to follow is PEMDAS to get it to the standard form Parentheses Exponent Multiplication Division Addition Subtraction [tex]y = \frac{1}{2} {(x - 4)}^{2} + 6 \\ y = \frac{1}{2}( (x - 4) \times (x - 4)) + 6 \\ y = \frac{1}{2} ( {x}^{2} - 4x - 4x + 16) + 6 \\ y = \frac{1}{2} ( {x}^{2} - 8x + 16) + 6 \\ y = \frac{1}{2}x - 4x + 8 + 6 \\ y = \frac{1}{2}x - 4x + 14[/tex] or option C in this case.
2) The quadratic parent function is [tex]y = {x}^{2} [/tex] in general if you want to move the parent function to move to the left r units you will have [tex]y = {(x + r)}^{2} [/tex] so in this case you will have [tex]y = {(x + 2)}^{2} [/tex] 3)
The equation to find the vertex is [tex]y = a {(x - h)}^{2} + k[/tex] since the paranola has vertex (2,-5) we have [tex]y = a {(x - 2)}^{2} - 5[/tex] to solve for a, we plug the values (3,4) into the equation and have the following
[tex]y = a {(x - 2)}^{2} - 5 \\ 4 = a {(3 - 2)}^{2} - 5 \\ 4 = a - 5 \\ 9 = a[/tex] once we have the value for a, we plus it into the previous equation to get the equation of a parabola with the desired vertex and that passes through the desired point so