How would I graph x^2-3x=4 to solve it?
And I would I check?
Thank You.

Start with the quadratic in the form y = a*x^2 + b*x + c.

If a>0, the parabola is U-shaped; if a<0, the parabola is upside-down u-shaped.

If a>0, the minimum y value is at the point called the vertex. (If a<0, the maximum
y value is at the point called the vertex.) The vertex (h,k) is found by
calculating h = -b/(2a), and then substituting that x value to find the corresponding
y value for k.

y = x^2 - 3x -4.

a=1, b=-3.

h = -b/(2a) = --3/(2*1) = +3/2

y = k = (3/2)^2 - 3*(3/2) -4 = -25/4.

So minimum y point is at (3/2, -25,4).

Also find the y-intercept where x=0: y = 0^2 - 3*0 -4 = -4.

Now just plug in various x values and graph. Try x=1, x=2, x=3, x=4, x=-1. x=-2. x=-3. x=-4.

3x^2-5x-2

Why are you posting on a topic which was posted almost a year and a half ago? How about starting your own thread?