<p>if the y coordinate is (0, -5) and two points that are (2, -3) and (-2, 9) fall on the quadratic graph function...what is the graph in standard form and what are their solutions?</p>
<p>If you have two points of a function, you can get the slope. You already have the value of the y-intercept.</p>
<p>(y2-y1)=m(x2-x1)</p>
<p>roflkeke, that's for a line. This clearly isn't a line.</p>
<p>xolala, I would try graphing and go from there.</p>
<p>Y=Ax^2+bx+c
you have three points and three unknowns, so set up a system of linear equations. plug in the point into x and y to get the three equations.</p>
<p>artofproblemsolving.com <--- go to their forums.</p>
<p>"roflkeke, that's for a line. This clearly isn't a line."</p>
<p>Whoops. When I saw it, I thought there was only one other point besides (0,-5).</p>
<p>do you have a graphing calculator?
if not then ill tell you how to do it by hand in a few minutes.</p>
<p>"f the y coordinate is (0, -5) and two points that are (2, -3) and (-2, 9) fall on the quadratic graph function...what is the graph in standard form and what are their solutions?"</p>
<p>plug the values into the formula y=ax^2+bx+c</p>
<p>you get the following equations:
-5= 0a + 0b + c
-3 = 4a + 2b + c
9 = 4a - 2b + c</p>
<p>Solve for a, b, and c. From there, plug the values of a, b, and c into the formula that I gave above (y=ax^2+bx+c). From there just turn that into the standard equation (complete the square)</p>