How to graph a parabola #y=x^2-4x-5#?

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Answer 1 · 2015-05-25T19:18:49

You can start by finding the x-intercepts, if there are any. That is when y = 0.

#0 = x^2 - 4x - 5#
#0 = (x - 5)(x + 1)#

So you have one intercept at #x = 5# and one at #x = -1#. Or, at #"(5, 0)"# and #"(-1, 0)"#.

Then, you can find where the vertex is (where the graph turns around). This can be found using:

#x = (-b)/(2a)#

#(-(-4))/(2*1) = 4/2 = 2#

Then you can plug it into the original equation and solve for the y-coordinate.

Plugging it in:
#x^2 - 4x - 5 = y#
#2^2 - 4*2 - 5 = 4 - 8 - 5 = -9#

So I would expect a vertex at #"(2, -9)"#.

If you want to go even further, you can solve to see if there are any y-intercepts (at #x = 0#).

#y = 0^2 - 4*0 - 5 = -5#

So there is also a y-intercept at #"(0, -5)"#.

Here's the graph:
graph{x^2 - 4x - 5 [-20.07, 20.06, -10.03, 10.04]}

You can click on the graphed curve to locate the intercepts and vertex.