using a graph what's the solution for x^2 + 4x - 5=0?

I find this a bit odd, because if you're given a really clean graph, it's easy to solve. Alternatively, if you're asked to graph a parabolic equation, you plot the vertex and typically you'll be asked to mark the focus and directrix as well. Then you do your best job at freehand which means your solution is only as good as your interpretive drawing.

Remember the solution for a vertical parabola is where the graph crosses the x-axis, where y=0. For example, if you were given a graph of a vertical parabola that crossed the x-axis at (-2,0) and (4,0), then it's solution is x=-2, 4.

Let's cut the difference a bit, but I think you should clarify on what you're being asked to present i.e. the expectation of accuracy.

GRAPH:  x^2+4x-5

Initial Assessment:

x^2 (vs y^2) so it's a vertical parabola, that is the line of symmetry is x=h

The coefficient of the x^2 term is positive, so the parabola will open upward.

vertex: (h,k)  

Convert the standard form ax^2+bx+c  to vertex form a(x-h)^2+k by completing the square

x^2+4x-5  => (x^2+4x+4) -5-4  => (x+2)^2 -9  Therefore the vertex is at (-2,-9)

Note, you don't have to change to vertex form. It's a matter of preference. If you look at the math you do to convert to vertex form, you can see that the vertex can also be defined as (-b/2a, c/a-(b/2a)^2).

So, plot a point (the vertex) at (-2,-9). Since it's an upward facing parabola, you know this is the lowest point and the parabola will indeed cross the x-axis in two points. But where?!

SOLVE: from the graph of x^2+4x-5 = (x+2)^2-9

So, here's the odd part, how wide or narrow do you draw the graph? You could calculate the focus (point inside the parabola) and the directrix (line outside the parabola perpendicular to the line of symmetry). By definition, a parabola is the set of all points that are equidistant from the focus and directrix. This makes it easier to draw, but it's still wildly objective, ehhh "creative".

Table: You can calculate some points, say one or two on each side of the line of symmetry. That should allow you to make a more accurate sketch. It should be obvious by now that unless we have a bunch of points, this will still be much closer to a "sketch" than a "graph".

The axis of symmetry is x= h= -2  so let's use the following table:

x           y

-2         -9    x = h (vertex)

-3,-1     -8    x = h ± 1

-4,0      -5    x = h ± 2

Now your sketch should be more accurate and perhaps you accurately guess that the solution (where the parabola crosses the x-axis) is x = -5, +1

Or perhaps you extend your table to x = h ± 3 and find it that way.

Or (so here's why I find this a bit odd), you look at (x+2)^2-9 and say well when (x+2) = ±sqrt(9), y=0. So, (x+2) = ±3  or x=-5,+1. But then, you've actually solved the problem algebraically just to plot the two x-intercepts. Which came first, the graph or the solution?

I think you may want to clarify the expectations for the method and accuracy of your solution to this problem. Perhaps they want you to use a graphing calculator, copy the graph, and point out the solution. It would be best to find out for sure.