Recall how the roots of quadratic functions can be found algebraically, using the quadratic formula $(x=\frac{-b \pm \sqrt {b^2-4ac}}{2a})$. The roots of a quadratic function can also be found graphically by making observations about its graph. These are two different methods that can be used to reach the same values, and we will now see how they are related.

Consider the quadratic function that is graphed below. Let's solve for its roots both graphically and algebraically.

Graph of the quadratic function $f(x) = x^2 - x - 2$

Graph showing the parabola on the Cartesian plane, including the points where it crosses the x-axis.

Notice that the parabola intersects the $x$-axis at two points: $(-1, 0)$ and $(2, 0)$. Recall that the $x$-intercepts of a parabola indicate the roots, or zeros, of the quadratic function. Therefore, there are roots at $x = -1$ and $x = 2$.

Now, let's solve for the roots of $f(x) = x^2 - x- 2$ algebraically with the quadratic formula.

Recall that the quadratic equation sets the quadratic expression equal to zero instead of $f(x)$:

$0 = x^2 - x - 2$

Now the quadratic formula can be applied to find the $x$-values for which this statement is true. For the given equation, we have the following coefficients: $a = 1$, $b = -1$, and $c = -2$.

Substitute these values in the quadratic formula:

$x = \dfrac{-(-1) \pm \sqrt {(-1)^2-4(1)(-2)}}{2(1)}$

Simplifying, we have:

$x = \dfrac{1 \pm \sqrt {9}}{2} \\$

and 

$x = \dfrac{1 \pm 3}{2}$

We now have two possible values for x: $\frac{1+3}{2}$ and $\frac{1-3}{2}$.

These reduce to $x = 2$ and $x = - 1$, respectively. Notice that these are the same values that when found when we solved for roots graphically.

Example

Find the roots of the quadratic function $f(x) = x^2 - 4x + 4$. Solve graphically and algebraically.

The graph of $f(x) = x^2 - 4x + 4$.

The graph of the above function, with the vertex labeled at $(2, 1)$.

Looking at the graph of the function, we notice that it does not intersect the $x$-axis. Therefore, it has no real roots.

We can verify this algebraically. First, identify the values for the coefficients: $a = 1$, $b = - 4$, and $c = 5$.

Substituting these into the quadratic formula, we have:

$x=\dfrac{-(-4) \pm \sqrt {(-4)^2-4(1)(5)}}{2(1)}$

Simplifying, we have:

$x=\dfrac{4 \pm \sqrt {16-20}}{2} \\ x=\dfrac{4 \pm \sqrt {-4}}{2}$

Notice that we have $\sqrt{-4}$ in the formula, which is not a real number. Therefore, there are no real roots for the given quadratic function. We have arrived at the same conclusion that we reached graphically.