Given function f(x)=x2+4x-5 determine the vertex, axis of symmetry calculate y-intercept find additional point on graph graph function. Submit graph.Need answer asap.
First thing to notice is that we're dealing with a parabola, so we know it has a vertex, vertical axis of symmetry, etc.
Always factor if possible.`f(x) = x^2+4x-5=(x-1)(x+5)`
Start by finding the roots. If 0 = (x-1)(x+5), then x = 1 or -5.
Now hit up the y-intercept. If x = 0, `f(0)=-5` .
The vertex is going to be right between the roots, so average them: `(1+(-5))/2)=-2`. The axis of symmetry is x -2.
So the vertex is at x = -2, and y = `f(-2)=(-2)^2+4(-2)-5=4-8-5=-9` . (-2, -9).
To find another point on the graph, you just choose a random x. How about 3. `f(3)=3^2+4*3-5=9+12-5=16` . (3, 16)
Double-check this graph to make sure the work was correct...