How Do You Find Tangent Lines to a Parabola Through a Given Point?

AI Thread Summary
To find the tangent lines to the parabola y = x^2 + x that pass through the point (2,-3), first calculate the derivative, which is 2x + 1. Substituting x = 2 into the derivative gives a slope of 5. Using the point-slope form of a line, the equation of the tangent line is derived as y = 5x - 10 - 3, simplifying to y = 5x - 13. This process illustrates the relationship between derivatives and tangent lines, emphasizing practice in differentiation for better understanding.
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hi,
i know this problem has to do with derivatives and I'm sure its easy, but for some reason i just can't figure it out...
Find equations of both lines through the point (2,-3) that are tangent to the parabola y= x^2 + x

help please...i know the derivative is 2x + 1 ...and that the 2 equations will be linear and pass through the point above, obviously...just not sure where to go from there. :blushing:
 
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This is differentiation, you have dy/dx =2x+1.then substitute the x=2 in the dy/dx,then you will find that dy/dx =5(that is the gradient of the tangent)


y-(-3) =3(x-2)
y+3 = 3x-6
y= 3x-9 (equation of the tangent)

The othe equation you mention may be the normal for the curve.Use formula

m1 x m2 =-1

then you can solve the question.
 


Hi there,

Don't worry, derivatives can be tricky at first but with practice, you'll get the hang of it! Let's break down this problem step by step.

First, you're correct that the derivative of y = x^2 + x is 2x + 1. This is because the derivative of x^2 is 2x, and the derivative of x is 1.

Next, we want to find the equations of lines that are tangent to the parabola at the point (2,-3). This means that the lines will have the same slope as the parabola at that point.

To find the slope of the parabola at (2,-3), we can plug in x = 2 into the derivative 2x + 1. This gives us a slope of 5. So, the lines we are looking for will have a slope of 5.

Now, we can use the point-slope form of a line to find the equations. The point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point. Plugging in m = 5 and (x1, y1) = (2,-3), we get:

y - (-3) = 5(x - 2) and y + 3 = 5x - 10

These are the equations of the two lines that are tangent to the parabola at (2,-3). You can simplify them if you'd like to get the final equations in slope-intercept form (y = mx + b).

I hope this helps! Don't be discouraged if you struggle with derivatives at first, just keep practicing and it will become easier. Good luck!
 
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