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Locating Intercepts of Parabola
Earlier, we learned how to sketch the graph of a parabola f x = a x − h + k
2 ( ) ( ) :
- Plot the vertex ( h , k ).
- If a > 0 , the parabola faces up; if a < 0 , the parabola faces down.
- If | a |> 1 , make the parabola thinner than the standard parabola
2 y = x ; if | a |< 1 , make the
parabola wider than
2 y = x ; if | a |= 1 , leave the parabola as wide as
2 y = x.
In the last lesson, we learned how to locate the vertex even if the function is given in standard form
g x = ax + bx + c
2 ( ). However, there is still an issue: If | a |≠ 1 , we don't know exactly how much
thinner or wider the graph should become.
In this lesson, we will learn how to make our sketched parabolas more accurate by locating its y -
intercept and x -intercept(s), if any.
Locating Parabola's Y-Intercept
Let's review how to find the y -intercept of y = 2 x + 3. I hope you didn't simply memorized that the "3"
is the y -value of the line's y -intercept.
Since a function's y -intercept crosses the y -axis, its x value must be 0―otherwise the point would not be
on the y -axis. This is why, to find a function's y-intercept, we plug x = 0 into the equation and then
solve for y. For y = 2 x + 3 , we have:
y = 2 ⋅ 0 + 3 = 3
This is why the y -intercept of y = 2 x + 3 is ( 0 , 3 ).
We apply the same rule to find a quadratic function's y -intercept.
[ Example 1 ] Find the y -intercept of ( ) 3 2 10
2 p x = x − x + and ( ) 3 ( 2 ) 10
2 q x = x − +.
[ Solution ] To find a function's y -intercept, we plug in x = 0. We have:
2 p = ⋅ − ⋅ + = , and
2 2 q = − + = ⋅ + =.
Solution: The y -intercept of p ( x )is ( 0 , 10 ), and the y -intercept of q ( x )is ( 0 , 22 ).
Note that any function can have at most one y -intercept. If a function has two y -intercepts, say ( 0 , 1 )
and ( 0 , 2 ), it implies the function has two output values when the input is 0, and that would disqualify
the relationship as a function.
For a parabola, it always has one and only one y -intercept.
Locating a Parabola's X-Intercept(s)
A parabola could have two, one or no x -intercepts. See the graph below:
Figure 1: parabolas have different number of x-intercepts
In this graph, the parabola at the top has no x -intercepts; the one in the middle has one x -intercept:
( 2 , 0 ); the one at the bottom has two x -intercepts: ( 1 , 0 )and ( 3 , 0 ).
Since a function's x -intercept is always on the x -axis, its y -value must be 0. We can clearly see this in
Figure 1. The strategy to find a function's x -intercept is to plug in y = 0 and then solve for x.
In the following examples, we will try to find the x -intercept(s) of those 3 functions in Figure 1 (pretend
you don't know them yet).
The left side of this equation can be factored, so we have:
1 or 3
1 0 or 3 0
x x
x x
x x
There are two solutions, x = 1 and x = 3 , for 4 3 0
2 x − x + = , implying the function
2 h x = x − x + has two x -intercept: (1, 0) and (3, 0). Figure 1 verifies this result.
Earlier, we learned that a quadratic equation's discriminant, b 4 ac
2 − , determines the equation has 2, 1
or 0 solutions.
In this lesson, we just learned that a quadratic function can have 2, 1 or 0 x -intercept.
Is there a relationship between these two observations? Let's look at this table.
Example 2 Example 3 Example 4
quadratic function (^) f ( x )= x^2 − 4 x + 5 g ( x )= x^2 − 4 x + 4 h ( x )= x^2 − 4 x + 3
quadratic equation (^) x^2 − 4 x + 5 = 0 x^2^ − 4 x + 4 = 0 x^2 − 4 x + 3 = 0
value of discriminant (^) b^2 − 4 ac =− 4 b^2 − 4 ac = 0 b^2 − 4 ac = 4
solutions none (^) x = 2 x = 1 and x = 3
x-intercepts none (2, 0) (1, 0) and (3, 0)
Please take time to understand the following important relationships:
- When we solve a quadratic equation 0
2 ax + bx + c = , we are actually looking for the x -
intercepts of the quadratic function q x = ax + bx + c
2 ( ).
2 b − ac > , the equation 0
2 ax + bx + c = has two solutions, and the quadratic
function q x = ax + bx + c
2 ( ) has two x -intercepts. This is the case for h ( x )in Figure 1.
2 b − ac = , the equation 0
2 ax + bx + c = has one solution, and the quadratic
function q x = ax + bx + c
2 ( ) has one x -intercept. The graph of q ( x )touches the x -axis at one
point, like g ( x )in Figure 1.
2 b − ac < , the equation 0
2 ax + bx + c = has no real solution, and the quadratic
function q x = ax + bx + c
2 ( ) has not x -intercept. The whole graph of q ( x )is either above or
below the x -axis, like f ( x )in Figure 1.
Studying mathematics is about understanding these relationships.
Let's look at one more example, where the parabola is given in vertex form.
[ Example 5 ] Find the x -intercept(s) of ( ) 2 ( 3 ) 8
2 f x = x + −.
[ Solution ] To find f ( x )'s x -intercepts, plug in f ( x )= 0 , and we have:
2
2
x
f x x
It's not wise to multiply out
2 ( x + 3 ) and then deal with a big mess. Instead, we will use the square root
property:
2
2
2
2
x
x
x
x
Since 4 > 0 , we can square root both sides and have:
1 or 5
3 2 or 3 2
x x
x x
x
x
Solution : ( ) 2 ( 3 ) 8
2 f x = x + − 's x -intercepts are (− 1 , 0 )and (− 5 , 0 ).
Look at the step
2 4 = ( x + 3 ). In some other problem, at this step, you would see something like
2 − 4 =( x + 3 ). Since
2 ( x + 3 ) cannot generate a negative number, at this step, we can claim that the
parabola has no x -intercepts.
