equations of tanget lines using a point not on the function

[mr0everywhere]

equations of tanget lines using a point not on the function[solved]

find equations of the 2 tangent lines to the graph of f that pass through the indicated point.
the equation is f(x)=x^2 and the point is (1,-3), i know f'(x)=2x , i have also found that the slope of the lines would be (-3-y)/(1-x).
part i was stuck on:
2x=(-3-f(x))/(1-x) or 2x = (-3x-x²)/(1-x)
all tangent lines must have the equation f(x₂)=(2x₁)x₂+c, f(x₂)=-3, x₂=1. This gives me -3=(2x)1-x^2 or -x^2+2x+3=0. factor this out -(x-3)(x+1)=0. plug these in as x₁, and i get the equations y=6x-9 and y=-2x-1

 

[pka]

find equations of the 2 tangent lines to the graph of f that pass through the indicated point. the equation is f(x)=x^2 and the point is (1,-3), i know f'(x)=2x , i have also found that the slope of the lines would be (-3-y)/(1-x).

If I were you, I would solve \(\displaystyle \dfrac{x^2-(-3)}{x-1}=2x\) (slope=slope).

 

[Ishuda]

find equations of the 2 tangent lines to the graph of f that pass through the indicated point.
the equation is f(x)=x^2 and the point is (1,-3), i know f'(x)=2x , i have also found that the slope of the lines would be (-3-y)/(1-x). i am stuck at this point

The form of a line, y(x), going through a particular point (x₀, y₀) and having a given slope m₀ is
y = y₀ + m₀ (x - x₀)
The form of a tangent line, t(x), going a particular point (x₁, y₁) and having a slope m₁ is
t = y₁ + m₁ (x - x₁)
where
m₁ = f'(x₁)
where f' indicates the derivative of f; that is the slope of the tangent line is the value of the derivative at the point on the curve. Or, since y₁ = x₁², we have
t = x₁² + 2 x₁ (x - x₁)

Lines y and t are the same, (x₀, y₀) = (1, -3) and, since y is the tangent line line t, m₀ = m₁. So, set y equal to t and solve for x₁.