Question

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`?

Answer 1

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 = -b/"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 = -(-30/"2*5") = 3`
Plugging in `x` gives `y = 5*3^2 - 30*3 + 31 = -14`
Vertex location = `(3,-14)`

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*0^2 - 30*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 `(6,31)`. 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.