f(x) = 2x^4-6x^2+7

Question

 Find an equation of the line tangent to the graph of f at x=3

Answer 1

f(x) = 2x^4 - 6x^2 + 7

take the derivative to find slope

f'(x) = 8x^3 -12x = slope of the tangent line

=8(3)^3 -12(3) = 8(27)-36 = 216 -36 = 180 = slope at the point where x=3

when x = 3 y = f(3) = 2(3)^4 -6(3)^2 + 7 = 2(81) -6(9)+7 = 162 -54 +7 = 108+7 = 115

at the point (3,115) the tangent line has the equation: 180 = (y-115)/(x-3)

cross multiply

y-115 = (180)(x-3)

y-115 = 180x -540

y = 180x -540 +115

y = 180x - 425