The general form for a quadratic equation is y=ax^2+bx+c In this question, the equation you wish to graph should be written as y=-x^2+14x-49.
The value of a is called the stretch factor and in your case this is -1 which tells me two things, your parabola will look exactly the same as the basic parabola y=x^2 and it will be reflected (opens downward) on the x-axis.
To find the line of symmetry, I use the formula x= (-b)/(2a). b in you equation is b =-14 and a=-1. So x=(-14)/((2)(-1)). Simplify to get x=7. I now know that the line of symmetry passes through the vertex when x=7.
Substitute this value into the equation we are graphing and I get y=-(7)^2+14(7)-49. This simplifies to y=0 So the vertex of your parabola is (7,0) Plot this first important point on the grid.
Finally, because I know the "stretch factor" for this parabola is -1, I can now plot four other important points on the grid. The first two are (6,-1) and ( (8,-1) The other two are (5,-4) and (9,-4) Plot these four points and carefully sketch out the parabola passing through all five important points.
So, how did I come up with those four points. Knowing that the line of symmetry cuts the x-axis at 7, I simply filled in 5, 6 8, and 9 into the equation for x and calculated the values for y. Actually I don't have to calculate all four. Because of symmetry, x=6 and x=8 have to have the same value and x=5 and x=9 are also the same.