How do you graph #y=x^2-2x-5#?

Instead, you can also graph this by finding the roots. This is done by using the quadratic formula, ie #(-b+-sqrt(b^2 - 4ac))/(2a)#

The result of this is that the roots are at #x=-1.449# and #x=3.449#

The turning point will be at the point where the derivative is equal to zero. For this step, #y=x^2 - 2x - 5#

So #dy/dx = 2x - 2#

#dy/dx = 2x - 2 = 0#

#2x = 2#

#x = 1#

Finally, check the corresponding y-coordinate at #x=1#

#y=x^2 - 2x - 5#

#y = 1^2 - 2 - 5 = 1-7 = (-6)#

And as this is a quadratic, you just fill in the line in the usual #x^2# shape, making the line fit the points we have found above.

You can find the roots by using the quadratic formula:

#(-b+-sqrt(b^2-4ac))/(2a)#

Here,

#a=1#

#b=-2#

#c=-5#

Plug in.

#(-(-2)+-sqrt((-2)^2-4*1*-5))/(2*1)=>#

#(2+-sqrt(24))/2#

Simplify:

#(2+-sqrt(6*4))/2#

#(2+-2sqrt(6))/2#

#1+-sqrt(6)#

The y-int of the equation #ax^2+bx+c# is #c#.

The y-int here is #-5#.

To find the vertex, turn to vertex form by completing the square:

#y=a(x-h)+k# with #(h,k)# as the vertex:

#(x^2-2x+(1)^2-(1)^2)-5#

#(x-1)^2-6#

The vertex is #(1,-6)#

Here is a graph for reference: graph{x^2-2x-5 [-9, 11, -6.6, 3.4]}