How to graph y=x^2-2x-5?

Question

j2lie · Answer

@whpalmer4

whpalmer4 · Answer

Same way you graph nearly any function.  Pick some values of $x$, calculate the corresponding values of $y$, plot the points on your graph, draw a smooth curve through them.

I'd suggest $x = -3$ through $x = 5$ would be an appropriate range

whpalmer4 · Answer

I say that because of the position of the vertex.

j2lie · Answer

i tried but they don't work. it's very frustrating.

anonymous · Answer

try using a graphing calculator if your stuck

whpalmer4 · Answer

sure it does.  x = -3, y = (-3)^2 - 2(-3) -5 = 9+6-5 = 10
x = -2, y = (-2)^2 - 2(-2) -5 = 4 + 4 - 5 = 3
x = -1, y = (-1)^2 - 2(-1) - 5 = 1 + 2 - 5 = -2
x = 0, y = (0)^2 - 2(0) - 5 = -5
x = 1, y = (1)^2 - 2(1) - 5 = 1-2-5 = -6
x = 2, y = (2)^2 - 2(2) - 5 = 4 - 4 - 5 = -5
x = 3, y = (3)^2 - 2(3) - 5 = 9 - 6 - 5 = -2
x = 4, y = (4)^2 - 2(4) - 5 = 16 - 8 - 5 = 3
x = 5, y = (5)^2 - 2(5) - 5 = 25 - 10 - 5 = 10

plot those points.

j2lie · Answer

you're a genius. thank you.

j2lie · Answer

I just only need five.