How do you find the vertex and intercepts for (x+3)^2=16(y-2)(x+3)2=16(y2)?

1 Answer
Aug 20, 2017

Vertex ->(x,y)=(-3,2)(x,y)=(3,2)

y-intercpt ->(0,41/16)(0,4116)

x-intercept -> None

Explanation:

This is almost in 'completed square' form.

Sometimes called vertex form.

We need to get the yy on its own.

y-2=1/16(x+3)^2y2=116(x+3)2

y=1/16(xcolor(red)(+3))^2color(blue)(+2) larr" Now it is in vertex form"y=116(x+3)2+2 Now it is in vertex form
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color(green)("To determine the vertex")To determine the vertex

x_("vertex")=(-1)xx(color(red)(+3)) = -3xvertex=(1)×(+3)=3

y_("vertex") =color(blue)(+2)yvertex=+2

Vertex ->(x,y)=(-3,2)(x,y)=(3,2)
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color(green)("To determine the " y" intercept")To determine the y intercept

This occurs at x=0x=0 so by substitution:

y=1/16(x+3)^2+2" "->" "y=9/16+2 =41/16y=116(x+3)2+2 y=916+2=4116

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color(green)("To determine the " x" intercept")To determine the x intercept

This occurs at y=0y=0 so by substitution:

y=1/16(x+3)^2+2" "->" "-2=1/16(x+3)^2y=116(x+3)2+2 2=116(x+3)2

color(white)("mmmmmmmmmmmb")->" "-32=(x+3)^2mmmmmmmmmmmb 32=(x+3)2

Square root both sides

sqrt(-32)=x+3 color(red)(larr" This result has a problem")32=x+3 This result has a problem

You can take the square root of a negative number but the answer is in the realm of Complex Numbers. Consequently the graph does not cross the x-axis

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Foot Note: The x^2x2 term is positive so the graph is of shape uu.

The vertex (which is a minimum value in this case) is above the x-axis. That alone tells you that there is no solution to y=0y=0 that is in the set of 'Real Numbers'. That is y!=0" for " x inRR

Tony BTony B