Find an equation of the line tangent to the graph of x^2+(y-x)^3=9 at x=1
you can see that graphs at
www.desmos.com/calculator/kwih0zk661
x^2+(y-x)^3=9
at x=1, y=3
2x + 3(x-y)^2 (y' - 1) = 0
y' = 1 - 2x/(3(y-x)^2)
at (1,3) y' = 1 - 2/12 = 5/6
so the tangent line is y-3 = 5/6 (x-1)
Thank you
self
To find the equation of the line tangent to the graph of the equation at a specific point, you need to find the slope of the tangent line and the coordinates of the point.
Step 1: Find the slope of the tangent line
To find the slope of the tangent line, you can differentiate the equation implicitly with respect to x. Let's differentiate the equation x^2 + (y - x)^3 = 9:
d/dx [x^2 + (y - x)^3] = d/dx [9]
Differentiating the equation gives us:
2x + 3(y - x)^2 * (1 - 1) = 0
Simplifying this equation gives us:
2x + 3(y - x)^2 = 0
Step 2: Find the coordinates of the point
To find the coordinates of the point, substitute x = 1 into the original equation:
(1)^2 + (y - 1)^3 = 9
Simplifying this equation gives us:
1 + (y - 1)^3 = 9
(y - 1)^3 = 8
Taking the cube root of both sides:
y - 1 = 2
y = 3
So, the coordinates of the point are (1, 3).
Step 3: Write the equation of the tangent line
Now that we have the slope of the tangent line (2x + 3(y - x)^2 = 0) and the point (1, 3), we can use the point-slope form of a linear equation to write the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the slope (m = 2x + 3(y - x)^2) and the point (x1 = 1, y1 = 3) into the equation, we get:
y - 3 = (2(1) + 3(y - 1)^2)(x - 1)
Simplifying this equation gives us the equation of the line tangent to the graph of x^2 + (y - x)^3 = 9 at x = 1:
y = 3(x - 1) + 3