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Find the vertex of the graph of the equation $x - y^2 + 8y = 13.$

Find the vertex of the graph of the equation $y = -2x^2 + 8x - 15.$

Find the area of the region enclosed by the graph of $x^2 + y^2 = 2x - 6y + 6$

The graph of $y = ax^2 + bx + c$ has an axis of symmetry of $x = 4.$ Find $\frac{b}{a}$.

The graph of $y = ax^2 + bx + c$ is shown below. Find $a \cdot b \cdot c$.  (The distance between the grid lines is one unit.)

May 28, 2021

#2
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1. The vertex of x - y^2 + 8y = 13 is (-8,4).

2. The vertex of y = -2x^2 + 8x - 15 is (4,-15).

May 28, 2021
#3
+34340
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1.)  x = y^2 -8y + 13

re-arrange to vertex form

x = (y-4)^2  -3        vertex is   -3,4

May 28, 2021
#4
+34340
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2 )   -2x^2+8x-15     re-arrange to vertex form

-2 (x^2 - 4x) -15

-2 ( x-2)^2  - 7         vertex = 2, -7

May 28, 2021
#5
+34340
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3)  x^2 -2x    + y^2  + 6y   =  6     Arrange into standard circle equation form

(x-1)^2      +      ( y+3)^2 = 6 +1 + 9

6+ 1 + 9 = 16  = r^2

Area of a circle = pi r^2 = pi (16)  = 16 pi  units2

May 28, 2021
#6
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Can you answer the last two questions? Thank you so much for the other answers.... all correct!

Jun 5, 2021