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

Question

Eli Assouline · Accepted Answer

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.

Eva Abramson · Answer

undefined To graph the parabola $y = x^2 - 4x - 5$, follow these steps:

1. Find the vertex of the parabola using the formula $x = -\frac{b}{2a}$.
2. Substitute the $x$-coordinate of the vertex into the equation to find the $y$-coordinate.
3. Plot the vertex on the coordinate plane.
4. Find additional points by choosing values of $x$ and calculating the corresponding values of $y$ using the equation.
5. Plot the additional points and draw the parabola through them.
6. Label the vertex and any other important points on the graph.