2007-10-19 15:35:48 · 4 answers · asked by ausomadam 1 in Science & Mathematics ➔ Mathematics
y=x^2+4x-5
y=(x+1)(x-5)
(-3,16) (-2,7) (-1,0) (0,-5) (1,-8) (2,-9) (3,-8) (4,-5) (5,0) (6,7) (7,16)
Graph the rest; I cant do it all for you; GOOD LUCK!
2007-10-19 15:40:09 · answer #1 · answered by Sanchez_150 2 · 0⤊ 0⤋
Graph the parabola.
It opens up, with a vertex of (-2, -9) and y intercept of (0, -5).
The solutions are where it crosses the x axis ( x=-5 and x = 1).
2007-10-19 15:49:05 · answer #2 · answered by RickSus R 5 · 0⤊ 0⤋
Get busy and make up a table of x and f(x) value pairs and then get them on a piece of graph paper.
Doug
2007-10-19 15:42:27 · answer #3 · answered by doug_donaghue 7 · 0⤊ 1⤋
use the quadratic equation
(-b plus or minus the sqrt(b^2 - 4ac)) / 2
2007-10-19 15:40:37 · answer #4 · answered by wicky034 2 · 0⤊ 0⤋