Answer:
The Graph of a Quadratic Function
A quadratic function is a polynomial function of degree 2 which can be written in the general form.
Step-by-step explanation:
Example: 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)f(0)===−x2−2x+3−(0)2−2(0)+33
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)000x+3x======−x2−2x+3−x2−2x+3x2+2x−3(x+3)(x−1)0−3orx−1x==Set f(x)=0.Multiply both sides by −1.Factor.Set each factor equal to zero.01
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=−b2a 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====−b2a−(−2)2(−1)2−2−1
Substitute −1 into the original function to find the corresponding y-value.
f(x)f(−1)====−x2−2x+3−(−1)2−2(−1)+3−1+2+34
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 −2 y 3f(−2)=−(−2)2−2(−2)+3=−4+4+3=3Point(−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:x−intercepts:Vertex:Extra point:(0,3)(−3,0) and (1,0)(−1,4)(−2,3)
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