How do you graph #y= (x+1)^2 -10 #?

1 Answer
Truong-Son N.
Aug 6, 2015

Three points make a curve. You can solve for the vertex (minimum) and the (presumably) two x-intercepts, and sketch whatever connects all three points.

Normally you might do this:
#x = -b/(2a)#

But this form is easier. You basically have an #x^2# curve.

This one is shifted left 1 unit and down 10 units from #(0,0)#, because #x+1# in parentheses indicates a shift opposite to the sign (#+# is left, #-# is right), and the #10# outside of the parentheses corresponds to #+# as up and #-# as down.

So instead of the vertex at #(0,0)#, it is at #color(blue)(((-1,-10)))#. The x-intercepts, you get from solving the equation set to #y = 0#:

#0 = (x+1)^2 - 10#
#pmsqrt10 = x+1#

#color(blue)(x_"right") = sqrt10 - 1 color(blue)(~~ 2.162)#
#color(blue)(x_"left") = -sqrt10 - 1 color(blue)(~~ -4.162)#

Finally, if you want to, you get the y-intercept by setting #x = 0#:
#y = (0 + 1)^2 - 10 = -9#, and it is #color(blue)(((0, -9)))#.

You can see that here by clicking on the intercepts and the vertex:

graph{(x+1)^2 - 10 [-5, 5, -15, 5]}

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