contestada

What is the standard form of the equation y = 1/2 (x - 4)^2 + 6
A) y= 1/2 x^2 − 8x + 22
B) y= 1/2 x^2 − 4x + 22
C) y= 1/2 x^2 − 4x + 14
D) y= 1/2 x^2 − 8x + 14

Which 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 + 2

Find 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)^2−5
y=2(x−2)^2−5
y=.75(x−2)^2−5
y=9(x−2)^2−5

Respuesta :

npichard
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

[tex]y = 9 {(x - 2)}^{2} - 5[/tex]