Write an equation for a tangent line to the graph of y=f(x) at the point on the graph where x has the indicated value. F(x) = 10x^2-6/4x-2, x=0
To find the equation of the tangent line to the graph of y = f(x) at the point where x = 0, we need to find the derivative of f(x) and evaluate it at x = 0.
Given f(x) = (10x^2 - 6) / (4x - 2), we can find the derivative f'(x) by applying the quotient rule:
f'(x) = [(10x^2 - 6)'(4x - 2) - (10x^2 - 6)(4x - 2)'] / (4x - 2)^2
= [(20x)(4x - 2) - (10x^2 - 6)(4)] / (4x - 2)^2
= (80x^2 - 40x - 40x^2 + 24 - 40x + 24) / (4x - 2)^2
= (40x^2 - 80x + 48) / (4x - 2)^2
= (40(x^2 - 2x + 1.2)) / (4x - 2)^2
= (10(x^2 - 2x + 1.2)) / (x - 0.5)^2
Now, to find the slope of the tangent line at x = 0, we plug in x = 0 into f'(x):
m = f'(0) = (10(0^2 - 2(0) + 1.2)) / (0 - 0.5)^2 = (10(1.2)) / 0.5^2 = 12 / 0.25 = 48
Therefore, the slope of the tangent line at x = 0 is 48. The equation of the tangent line, using the point-slope form, is:
y - f(0) = m(x - 0)
y - f(0) = 48x
Substituting f(0) by plugging x = 0 into f(x), we have:
y - (10(0^2) - 6) / (4(0) - 2) = 48x
y + 6 = 48x
Thus, the equation of the tangent line to the graph of y = f(x) at the point where x = 0 is y + 6 = 48x.