Find the axis of symmetry for this parabola
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Answer:
Step-by-step explanation:
Remember the general form of a linear equartion: y=mx + b, where m is the slope and b is the y-intercept (the value of y when x = 0).
The equation of line v (y=9x+1) has a slope of 9. A perpendicular line with have a slope equal to the negative inverse of the reference line. So the line we want will have a slope of -(1/9). That gives us y = -(1/9)x + b. Any vale of b will produce a line that is perpendicular. But we want a line that goes through point (3,-2). All we need to do is find a value for b that will make that wish come true.
Use the point (3,-2) in y = -(1/9)x + b and solve for b:
y = -(1/9)x + b
-2 = -(1/9)*(3) + b
-2 = -(1/3) + b
b = - 1 2/3, or -(5/3)
The equation is y = -(1/9)x + - (5/3)
