hi, i'd really appreciate some help solving this question! I am completely at loss here :/ The question is to:
Find the equation of the normal to y=loge(x+2) which is parallel to the line with equation y+3x-5=0
I am fine at solving these sorts of equations when they give me one equation and an provide a given x value but in this question i have no idea what i am meant to be doing... Please help!!
Put the second equation in slope intercept form
y=-3x +5 slope is -3
So what is the slope of the tangent to
y=ln(x+2)
slope= 1/(x+2)
so the inverse negative slope will be
- (x+2), this has to equal -3x
set them equal, and you have the x where it occurs.
Put that x into the log equation, and find y.
y=mx +b
You have y, m, x, solve for b.
To find the equation of the normal to the curve y = ln(x+2) that is parallel to the line y + 3x - 5 = 0, you need to follow these steps:
Step 1: Convert the given line equation to slope-intercept form (y = mx + b) where m is the slope and b is the y-intercept. In this case, the given line equation is y + 3x - 5 = 0. Rearranging the equation, we get y = -3x + 5. Now, we can see that the slope of this line is -3.
Step 2: Determine the slope of the tangent line to the curve y = ln(x+2). The slope of the tangent line can be found by taking the derivative of the function y = ln(x+2) with respect to x. The derivative of ln(x+2) is 1/(x+2).
Step 3: Since the normal line is parallel to the given line, the slopes of the normal line and the given line are equal. So, the slope of the normal line is also -3.
Step 4: Now that we have the slope of the normal line and a point on the line (x, y), we can find the equation of the normal line using the point-slope form y - y1 = m(x - x1). Let's consider a general point on the curve (x, y).
Using the slope-intercept form of the line equation, we have y = -3x + b, where b is the y-intercept we need to find.
Step 5: Substitute the coordinates of the general point on the curve (x, y) into the equation of the line y = -3x + b. We get y = -3x + b = ln(x+2).
Step 6: Solve this equation for x to find the x-coordinate where the normal line intersects the curve y = ln(x+2). Solve: -3x + b = ln(x+2) for x.
Step 7: Once you find the value of x in the above equation, substitute it back into the equation y = ln(x+2) to find the corresponding y-coordinate.
Step 8: Lastly, substitute the values of x and y into the equation y = -3x + b to solve for b, the y-intercept of the normal line.
Once you have the values of m (slope) and b (y-intercept), you can write down the equation of the normal line in slope-intercept form (y = mx + b).