# How do you use the important points to sketch the graph of -x^2 + 14x - 49?

Mar 16, 2016

graph{-x^2+14x-49 [-10, 10, -5, 5]}

#### Explanation:

The general form for a quadratic equation is $y = a {x}^{2} + b x + c$ In this question, the equation you wish to graph should be written as $y = - {x}^{2} + 14 x - 49$.

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 $x = \frac{- b}{2 a}$. $b$ in you equation is $b = - 14$ and $a = - 1$. So $x = \frac{- 14}{\left(2\right) \left(- 1\right)}$. 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 $y = - {\left(7\right)}^{2} + 14 \left(7\right) - 49$. This simplifies to $y = 0$ So the vertex of your parabola is $\left(7 , 0\right)$ 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 $\left(6 , - 1\right)$ and ( $\left(8 , - 1\right)$ The other two are $\left(5 , - 4\right)$ and $\left(9 , - 4\right)$ 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.