The Intercepts of a Parabola
Example 1
Find the y- and x-intercepts of the function: f(x) = x2 - 4x + 4
Solution
The y-intercept is the point where x = 0. It is the point (0, C).
For the given function, it is (0, 4).
To find the x-intercepts, replace f(x) with 0 and then solve for x.
| Original function.
Substitute 0 for f(x). |
f(x) = x2 - 4x + 4
0 = x2 - 4x + 4 |
To solve for x
| Factor.
Set each factor equal to 0.
Solve each equation. |
0
x - 2
x |
= (x - 2)(x + 2) = 0 or x - 2 = 0
= 2 or x = 2 |
We have one solution, x = 2, of multiplicity 2. This means the parabola
touches the x-axis only once.
So, the x-intercept of f(x) = x2 - 4x + 4 is (2, 0).
Example 2
Find the y- and x-intercepts of the function: f(x) = x2 + 9
Solution
The y-intercept is the point where x = 0. It is the point (0, C).
For the given function, it is (0, 9).
To find the x-intercepts, replace f(x) with 0 and then solve for x.
| Original function.
Substitute 0 for f(x). |
f(x) = x2 + 9
0 = x2 + 9 |
To solve for x:
| Subtract 9 from both sides.
|
-9 |
= x2 |
| Take the square root of each side.
|
 |
= x |
| Simplify. |
±3i |
= x |
Since the solution is two imaginary roots, the function f(x) = x2
+ 9 has
no x-intercepts. That is, it does not cross the x-axis.
Note:
Remember, when you take the square root
of each side of an equation you must
include both the positive and negative
square roots. That is why we need the ± in
= x.
Remember, taking the square root of a
negative number results in an imaginary
number, which we indicate with the letter i.
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