Use the graph of y=x^2+4x-5 what are the solution(s) to the equation. does this function have a max or a min?

Answer 1

the solutions to your equation will be the intercepts on the x-axis. that function will have a minimum because it is a parabola opening upwards (positive x^2 opens up). the minimum will be the coordinates at the bottom of your graph.

Answer 2

The graph is a parabola that opens up with a y intercept of (0, -5) and a vertex (minimum) at (-2, -9). The solutions are where it crosses the x axis. x = -5 and x = 1

Answer 3

y = (x+5)(x-1) so roots are x = -5 and x = 1.
The x^2 term is positive so the graph will show a minimum at x = -4/(2*1) = -2.
The min value will be 4 -8-5 = -9

Answer 4

use a graphing calculator to graph the equation, and you'll see that the solutions (aka- wherever the graph crosses the x-axis) are (1,0) and (-5, 0) the minimum is the lowest point of the graph, which in the case is the vertex, and it's point is (-2, -9)


i hope this helps! :)

Answer 5

y=x^2+4x-5
solve for the x-intercepts
(x+5)(x-1)
x=-5,1
y-intercept
y=0^2+4(0)-5=-5
minimum pt since the parabola opens upward
x=-b/2a=-(4)/2(1)=-2
f(-2)=(-2)^2+4(-2)-5=4-8-5=-9
P(-2,-9) is minimum

Answer 6

(x+5) (x-1)

has a min (because its positive (opens upward)) it will be at x = 2. (since thats halfway between 5 and -1)

Answer 7

y+5=x^2+4x

y+5-4x=x^2

square root of (y+5-4x)=x

i don't real get this question, but i tried