8A F11 Q3
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Question: a) Find the vertex, standard graphing form, and X-intercept for
b) Sketch the graph. Provide the focus and directrix.
Note: In this problem, what is referred to as standard graphing form is the vertex form, in case you search on the internet.
Foundations |
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1) What type of function are we asking you to graph (line, parabola, circle, etc.)? |
2) What is the process for transforming the function into the standard graphing form? |
3) After we have the standard graphing form how do you find the X-intercept, and vertex? |
4) Moving on to part b) How do we find a point on the graph? |
5) From the standard graphing form how do we obtain relevant information about the focus and directrix? |
Answers: |
1) The function is a parabola. Some of the hints: We are asked to find the vertex, and directrix. Also only one variable, of x and y, is squared. |
2) First we complete the square. Then we divide by the coefficient of x. |
3) To find the X-intercept, replace y with 0 and solve for x. Since the parabola is in standard graphing form, the vertex of is (h, k). |
4) To find a point, we can either use the symmetry of a parabola or plug in another value for x. |
5) From the equation , we use the equation to find p. P is both the distance from the vertex to the focus and the distance from the vertex to the directrix. |
Solution:
Step 1: |
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There are two ways to obtain the standard graphing form. |
Regardless of the method the first step is the same: subtract 2 from both sides to yield |
Method 1: |
Divide both sides by -3 to make the coefficient of , 1. This means |
Complete the square to get |
Multiply both sides by -3 so , and simplify the left side to yield |
Method 2: |
Instead of dividing by -3 we factor it out of the right hand side to get . |
Now we complete the square inside the parenthesis and add -3 to the left hand side resulting in |