Find the equation of a straight line that passes through a point (1,5) and tangent to curve y = x^3
f(x)=x^3
f'(x)=3x^2
m=f'(1)=3(1)^2=3
y=mx+b where y=5, m=3, x=1
5=3*1+b
b=2
so equation of straight line is
y=3x+2
The point (1,5) isn't on the graph of y^x^3. It is just a point on the tangent of the graph y = x^3. Sorry if I wasn't clear before.
y = x^3*
To find the equation of a straight line that passes through a point and is tangent to a curve, we need to find the slope of the curve at the given point.
Given the curve y = x^3, we can find the slope of the curve at any point by taking the derivative of the equation with respect to x.
Taking the derivative of y = x^3, we get:
dy/dx = 3x^2
Now, let's find the derivative at x = 1, since the line we want to find will be tangent to the curve at that point:
dy/dx = 3 * (1)^2 = 3
The slope of the curve at x = 1 is 3.
Now that we have the slope of the tangent line, we can use the point-slope form of a straight line to find the equation.
The point-slope form of a line is given by:
y - y1 = m(x - x1)
where (x1, y1) represents the point on the line and m represents the slope of the line.
Using the point (1,5) and the slope 3, we can plug in these values into the equation:
y - 5 = 3(x - 1)
Expanding the equation:
y - 5 = 3x - 3
Rearranging the equation to the slope-intercept form, y = mx + b, where b is the y-intercept:
y = 3x + 2
Therefore, the equation of the straight line that passes through the point (1,5) and is tangent to the curve y = x^3 is y = 3x + 2.