find an equation for all tangent lines drawn to the graph f(x)= .5x^2 - 6x + 30 from the point (7,8)

Question

find an equation for all tangent lines drawn to the graph f(x)= .5x^2 - 6x + 30 from the point (7,8)

Answer 1

f'(x) = x - 6; Y = 8 + (x - 6)(X - 7); 0.5x² - 6x + 30 = 8 + x² - 13x + 42; 0.5x² - 7x + 12= 0; x² - 14x + 24 = 0;

(x - 2)(x - 12) = 0; x = 2; or x = 12.

f'(2) = 2 - 6 = - 4; f(2) = 2 - 12 + 30 = 20; y = 20 - 4(x - 2); y = - 4x + 28

f'(12) = 12 - 6) = 6; f(12) = 72 - 72 + 30 = 30; y = 30 + 6(x - 12); y = 6x - 42

Answer 2

This question is posed under the topic precalculus, but likely you need some calculus to solve this problem, i.e. find the first derivative of the function in order to find a formula for slopes of tangent lines to the graph of the given function.

Let (a, y) represent the point of tangency (on the graph of the function. That point can be represented in terms of a as (a, 1/2 a² - 6a + 30). The slope of a line passing through that point and (7, 8) based on the formula for slope between two points is: (1/2 a² - 6a + 30 - 8 ) / (a - 7).

The derivative of f(x) is x - 6. So the expression above for slope at a point with x-coordinate "a" is a - 6. Those two expressions must be equal.

Set up and solve that equation for "a". There will be two solutions. Since "a" represents the x-coordinate of the POT, you could substitute that value(s) into f(x) to get the y-coordinate, but that is not necessary to find the equation of the tangent line(s).

The equation of the tangent line will be y = f'(a)(x-7) - 8. (using (7,8) in the point-slope formula.

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