1) An equation of the line contains points (7/9, 7) and (-7/9) is
2) Find the slope of the line tangent to the curve y=x^2 at the point (-0.6, 0.36) and find the corresponding equation of the tangent line
#1 I only see one point and a number.
#2 at the given point, the slope is -1.2
So, the line there is
y-0.36 = -1.2(x+0.6)
1) To find the equation of the line containing the points (7/9, 7) and (-7/9), we can use the point-slope form of a linear equation. The equation is given by:
y - y1 = m(x - x1),
where (x1, y1) is a point on the line and m is the slope.
First, let's find the slope (m) using the coordinates of the two points:
m = (y2 - y1) / (x2 - x1),
where (x1, y1) = (7/9, 7) and (x2, y2) = (-7/9, y).
Substituting the coordinates into the slope formula, we have:
m = (7 - 7) / (-7/9 - 7/9)
= 0 / (-14/9)
= 0.
Since the slope (m) is 0, the equation of the line is a horizontal line. Any x-value will have the same y-value, which is 7.
Therefore, the equation of the line is y = 7.
2) To find the slope of the tangent line to the curve y = x^2 at the point (-0.6, 0.36), we need to find the derivative of the function y = x^2 and evaluate it at x = -0.6.
The derivative of y = x^2 with respect to x is given by dy/dx = 2x.
Now, let's substitute x = -0.6 into the derivative to find the slope:
m = dy/dx = 2(-0.6)
= -1.2.
The slope of the tangent line at the point (-0.6, 0.36) is -1.2.
To find the equation of the tangent line, we can use the point-slope form as mentioned earlier:
y - y1 = m(x - x1),
where (x1, y1) = (-0.6, 0.36) and m = -1.2.
Substituting the values into the equation, we have:
y - 0.36 = -1.2(x - (-0.6))
= -1.2(x + 0.6)
= -1.2x - 0.72.
Simplifying, we get:
y = -1.2x - 0.72.
Therefore, the equation of the tangent line to the curve y = x^2 at the point (-0.6, 0.36) is y = -1.2x - 0.72.