Find the equation of the line that passes through the point (3,14) and is parallel to y=2/3x -6
2 answers:
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9514 1404 393
Answer:
y = -3x^2 +3x +6
Step-by-step explanation:
For roots p and q, the equation can be written as ...
y = a(x -p)(x -q)
The value of 'a' must be determined so that the product <em>apq</em> is equal to the y-intercept. One could say that the formula is ...
y = (y-intercept)/(pq)·(x -p)(x -q)
For your given values of p = 2, q = -1, y-intercept = 6, this becomes ...
y = 6/(2(-1))(x -2)(x +1)
y = -3(x^2 -x -2) . . . . . simplifying a bit
y = -3x^2 +3x +6
