Find an equation of the tangent line to the curve at the point (0, 1).
y = (1 + 3x)20
i got y' = 60(3x+1)^19
what do i do next?
at the point (0,1) x=0, so sub that in
y' = 60(1)^19 = 60
but (0,1) is a y-intercept of 1
so equation is y = 60x + 1
y' = 60(3x + 1)^19
y' = 60(3(0) + 1)^19
y' = 60(1)=60
Then use point-slope form
y - y1 = 60(x - x1)
y - 1 = 60(x - 0)
y - 1 = 60x
y = 60x + 1
To find the equation of the tangent line to the curve at the point (0, 1), you need to use the point-slope form of a linear equation, which is given by:
y - y₁ = m(x - x₁)
Where (x₁, y₁) represents the point on the curve, and "m" represents the slope of the tangent line at that point.
In this case, the point (0, 1) is given, so x₁ = 0 and y₁ = 1. The slope "m" can be obtained by evaluating the derivative you found, y' = 60(3x+1)^19, at x = 0.
Let's substitute these values into the point-slope form:
y - 1 = 60(3(0)+1)^19 (x - 0)
Simplifying this equation, we get:
y - 1 = 60(1)^19 x
Since 1 raised to any power is still 1, the equation becomes:
y - 1 = 60x
Finally, we can rewrite this equation in slope-intercept form, y = mx + b, where "m" is the coefficient of x (which is 60 in this case), and "b" is the y-intercept (which is -1 in this case). Therefore, the equation of the tangent line is:
y = 60x + 1