Question

How do you graph `f(x)=-x^2+2x+5` and identify the x intercepts, vertex?

Answer 1

You can find the roots using the quadratic equation and the vertex using the formula `-b/{2a}`.

Explanation:

Recall the general form of a quadratic function: `f(x)=ax^2+bx+c`.

VERTEX: The vertex has an `x`-coordinate of `-b/{2a}` (you can figure this out with calculus, or by factorising the general expression into turning point form). Once we know the `x`-coordinate, we can substitute into `f(x)` to find the `y`-coordinate. In this case, we end up with the coordinates `(1, 6)` for the vertex.

Y-INTERCEPT: The graph crosses the `y`-axis when `x=0`, so substitute `x=0` and we find `y=f(0)=c`. Thus, for a quadratic of the form `f(x)=ax^2+bx+c`, the `y`-intercept is always c. In this case, the `y`-intercept is 5.

X-INTERCEPT: To find the `x`-intercept, we have to solve the equation `ax^2+bx+c=0`. This has to be done through factorisation. By factorising the general quadratic form (Google it), we get the quadratic equation:

`x={-b+-sqrt(b^2-4ac)}/{2a}`

Substituting gives us

`x=1+-\sqrt(6)`

as our roots, or `x`-intercepts.

Alternatively, we could try factorising the quadratic manually, this is easier if the quadratic has integer roots (or rational roots). In this case, we can't (well, we can, but not in terms of rational numbers).

Together, the roots, the vertex, and the `y`-intercept give you 4-points. Which is more than enough to graph the quadratic freehand. This is what it looks like:

Also, since the leading coefficient is negative, you know that the quadratic must but an upside-down U.