Find all point(s) of intersection of the line y = 4x and the parabola y = x^2 - 2x + 9. a) (3, 12) b) (3, 24) c) (-3, -12) d) (-3, -24)

A) (3, 12)

Step-by-step explanation:

For such a problem, I like to use a graphing calculator. It shows the answer quickly without a lot of fuss.

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If you want to solve this analytically, set the difference in y-values equal to zero and solve the resulting quadratic in the usual way. This will give the x-value at which the y-values are equal. (After we find x, we still need to find y.)

... y - y = 0

... x² -2x +9 -4x = 0

... x² -6x +9 = 0 . . . . . . collect terms. Recognize this as a perfect square.

... (x -3)² = 0

... x = 3 is the solution to this

... y = 4x = 4·3 = 12

The point on each of the given curves is (x, y) = (3, 12). The line is tangent to the parabola there, so there is only one point of intersection.

(1/x + 1)^2 + 6 (1/x + 1) +5

Ifound this on google “that means that the graph of a function with direct variation has these features: it is a straight line: because the slope is the ratio y/x = k. it passes through the origin (0,0): because y = kx = k (0) = 0.” i hope this you for what your looking for