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.
