# How do you find the vertex, the y-intercept, and symmetric point, and use these to sketch the graph given y=5x^2-30x+31?

May 31, 2018

To do this we will need to perform some algebraic manipulation.

#### Explanation:

Vertex
To find the vertex, we need the $x$ and $y$ location. There is a handy formula (when the equation is in standard form) for the $x$-coordinate: $x = - \frac{b}{\text{2a}}$. To find the $y$-coordinate, we can plug in the $x$ value we got for the vertex into the original equation.

Doing this gives $x = - \left(- \frac{30}{\text{2*5}}\right) = 3$
Plugging in $x$ gives $y = 5 \cdot {3}^{2} - 30 \cdot 3 + 31 = - 14$
Vertex location = $\left(3 , - 14\right)$

Y-intercept
The $y$-intercept is when the function crosses the $y$-axis. We can think of this as when the value of $x$ is equal to $0$.
Doing this gives $y = 5 \cdot {0}^{2} - 30 \cdot 0 + 31 = 31$.
Y-intercept = $31$.

Symmetric Point
A quadratic equation such as this will produce a parabola, which is always symmetric about its vertex. This means that the symmetric point will be the vertex.

Graphing
Mark a point where the vertex is, then mark the point with the $y$-intercept, and use the concept of symmetry to draw a third point at $\left(6 , 31\right)$. Hint: Draw a dashed vertical line over the vertex, and show that this line is a line of symmetry.

After you have done this, it is reasonable to sketch the parabola.