Question

1. the graph of ( y = x^2 ) is reflected in the x - axis, compressed vertically by ( \frac{1}{2} ), then translated 2 units to the left and 1 unit down. write the equation of the transformed parabola.

1. write the equation of the parabola shown:

(there is a graph of a parabola with points (-6, 4) and (-5, 3) and the coordinate axes marked with -6, -5, -2, 0 on the x - axis and -2, 2, 4 on the y - axis)

1. write the mapping notation that transforms each point on ( y = x^2 ) to the point on ( y=-3x^2 + 1 ) and then make the table of values for both relations.

1. graph each parabola using mapping notation. include table of values for both parent and transformed parabola.

( y =-(x - 2)^2+3 )
(there is a coordinate grid with x - axis from -10 to 10 and y - axis from -10 to 10)

Explanation:

Problem 1

Step1: Reflect over x-axis

$y = -x^2$

Step2: Vertically compress by $\frac{1}{3}$

$y = -\frac{1}{3}x^2$

Step3: Shift left 2 units

$y = -\frac{1}{3}(x+2)^2$

Step4: Shift down 1 unit

$y = -\frac{1}{3}(x+2)^2 - 1$

Step1: Identify vertex form

Vertex $(-6,4)$, so $y=a(x+6)^2+4$

Step2: Solve for $a$ with $(-5,3)$

$3 = a(-5+6)^2+4 \implies 3=a+4 \implies a=-1$

Step3: Substitute $a$ into formula

$y = -(x+6)^2+4$

Step1: Define mapping notation

For $(x,y) \to (x, -3y+1)$

Step2: Create value table

$x$	$y=x^2$	$y=-3x^2+1$
$-1$	$1$	$-3(1)+1=-2$
$0$	$0$	$-3(0)+1=1$
$1$	$1$	$-3(1)+1=-2$
$2$	$4$	$-3(4)+1=-11$

Answer:

$y = -\frac{1}{3}(x+2)^2 - 1$

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Problem 2