How do you use the important points to sketch the graph of #x^2 - 5x - 6#?

1 Answer
Jul 19, 2018

Since this is a parabola, we know its general shape.

There's two ways I'd do this: vertex properties and factoring

To work with vertex properties, the first thing is the symmetry axis.

We know how to find this:
#x = -b/(2a) = 5/2 #
We can plug that into the equation to find the point of the vertex:
#(5/2)^2 - 5(5/2) - 6 = 25/4 - 25/2 - 6 = - 49/4 #

We can then use the y-intercept of the function by plugging in #x=0# to find the point (0,-6) on the function. We know that it is an upward parabola so we should pretty easily be able to sketch the graph.

We could also notice that the function factors to (x-6)(x+1), meaning it has zeroes at #x=-1, 6#. Since it's a positive parabola (since the coefficient of #x^2# is greater than zero), we could also sketch it like this.

Either way, we get something like this

graph{x^2-5x-6 [-3, 10, -20, 10]}