Respuesta :
Step-by-step explanation:
The Graph of a Quadratic Function
A quadratic function is a polynomial function of degree 2 which can be written in the general form,
f(x)=ax2+bx+c
Here a, b and c represent real numbers where a≠0. The squaring function f(x)=x2 is a quadratic function whose graph follows.
This general curved shape is called a parabola and is shared by the graphs of all quadratic functions. Note that the graph is indeed a function as it passes the vertical line test. Furthermore, the domain of this function consists of the set of all real numbers (−∞,∞) and the range consists of the set of nonnegative numbers [0,∞).
When graphing parabolas, we want to include certain special points in the graph. The y-intercept is the point where the graph intersects the y-axis. The x-intercepts are the points where the graph intersects the x-axis. The vertex is the point that defines the minimum or maximum of the graph. Lastly, the line of symmetry (also called the axis of symmetry) is the vertical line through the vertex, about which the parabola is symmetric.
For any parabola, we will find the vertex and y-intercept. In addition, if the x-intercepts exist, then we will want to determine those as well. Guessing at the x-values of these special points is not practical; therefore, we will develop techniques that will facilitate finding them. Many of these techniques will be used extensively as we progress in our study of algebra.
Given a quadratic function f(x)=ax2+bx+c, find the y-intercept by evaluating the function where x=0. In general, f(0)=a(0)2+b(0)+c=c, and we have
y-intercept (0,c)
Next, recall that the x-intercepts, if they exist, can be found by setting f(x)=0. Doing this, we have a2+bx+c=0, which has general solutions given by the quadratic formula, x=
−b±
√
b2−4ac
2a
. Therefore, the x-intercepts have this general form:
x-intercepts (
−b−
√
b2−4ac
2a
,0) and (
−b+
√
b2−4ac
2a
,0)
Using the fact that a parabola is symmetric, we can determine the vertical line of symmetry using the x-intercepts. To do this, we find the x-value midway between the x-intercepts by taking an average as follows:
x = (
−b−
√
b2−4ac
2a
+
−b+
√
b2−4ac
2a
) ÷2 = (
−b−
√
b2−4ac
−b+
√
b2−4ac
2a
)÷(
2
1
) =
−2b
2a
⋅
1
2
= −
b
2a
Therefore, the line of symmetry is the vertical line x=−
b
2a
. We can use the line of symmetry to find the the vertex.
Line of symmetry Vertex x=−
b
2a
(−
b
2a
,f(−
b
2a
))
Generally three points determine a parabola. However, in this section we will find five points so that we can get a better approximation of the general shape. The steps for graphing a parabola are outlined in the following example.
Example 1
Graph: f(x)=−x2−2x+3.
Solution:
Step 1: Determine the y-intercept. To do this, set x=0 and find f(0).
f(x) = −x2−2x+3 f(0) = −(0)2−2(0)+3 = 3
The y-intercept is (0,3).
Step 2: Determine the x-intercepts if any. To do this, set f(x)=0 and solve for x.
f(x) = −x2−2x+3 Set f(x)=0. 0 = −x2−2x+3 Multiply both sides by −1. 0 = x2+2x−3 Factor. 0 = (x+3)(x−1) Set each factor equal to zero. x+3 = 0 or x−1 = 0 x = −3 x = 1
Here where f(x)=0, we obtain two solutions. Hence, there are two x-intercepts, (−3,0) and (1,0).
Step 3: Determine the vertex. One way to do this is to first use x=−
b
2a
to find the x-value of the vertex and then substitute this value in the function to find the corresponding y-value. In this example, a=−1 and b=−2.
x =
−b
2a
=
−(−2)
2(−1)
=
2
−2
= −1
Substitute −1 into the original function to find the corresponding y-value.
f(x) = −x2−2x+3 f(−1) = −(−1)2−2(−1)+3 = −1+2+3 = 4
The vertex is (−1,4).
Step 4: Determine extra points so that we have at least five points to plot. Ensure a good sampling on either side of the line of symmetry. In this example, one other point will suffice. Choose x=−2 and find the corresponding y-value.
x y Point −2 3 f(−2)=−(−2)2−2(−2)+3=−4+4+3=3 (−2,3)
Our fifth point is (−2,3).
Step 5: Plot the points and sketch the graph. To recap, the points that we have found are
y-intercept: (0,3) x-intercepts: (−3,0) and (1,0) Vertex: (−1,4) Extra point: (−2,3)
Answer:
The parabola opens downward. In general, use the leading coefficient to determine if the parabola opens upward or downward. If the leading coefficient is negative, as in the previous example, then the parabola opens downward. If the leading coefficient is positive, then the parabola opens upward.
All quadratic functions of the form f(x)=ax2+bx+c have parabolic graphs with y-intercept (0,c). However, not all parabolas have x-intercepts.
