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)^2y−2=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 B