Equation of tangent

[mikeyjack21]

Hi,
Someone please help

Find the equation of the tangent line of y= 7e^(5x^2+7x-1) if the slope of the tangent is -21e^-3



the answer is y= -21e^-3x -14e^-3 but i am unable to find the x value

 

[lev888]

Hi,
Someone please help

Find the equation of the tangent line of y= 7e^(5x^2+7x-1) if the slope of the tangent is -21e^-3



the answer is y= -21e^-3x -14e^-3 but i am unable to find the x value

What's the relationship between a function's derivative and its graph's tangent lines?

 

[mikeyjack21]

the slope of the tangent line is the derivative of the function at some specific point.

 

[mikeyjack21]

This is what i have so far, i got 3 x values but when i sub them back in the equation, only x=-1 seems to be working. I do not understand why tho. Even looked at desmos and the tangent occurs at (-1, 0.3485)

 

[Steven G]

First of all do not distribute the 7. It is not wrong but possibly you might divide by a factor or multiple of 7. And you did when you divided by 21.

The log equation is not correct at all. Go and review your log rules. Actually they are.

ln(a) + ln(b) = ln(ab), ln(a) - ln(b) = ln(a/b) and ln(a^r) = rln(a). These are enough rules for you to handle your equation.

Also if a fraction equals 1, can you just set the numerator equal to the denominator!