secant equation

[Matt2015]

Hello, I am in an advanced functions course and have no experience with calc or any other courses. I am stuck on a question, I am not sure what to do or what is correct. Any help would be appreciated.

Determine the general equation of the secant from any point (A,f(A)) to the point on f(x) = 4^x where x = 2. Describe how you would use this to find the equation of the tangent at the point (2,f(2)).

well I know secant is y=m(x-p)+q where m = slope = y2-y1/x2-x1

The general equation for the secant would be something like (f(A)-16 / A-2)x+16

Is this correct?

The second part of the equation asks how I would use this to find the equation for the tangent at the point (2,f(2)).
What would I do here? I feel like its repeating the question.

 

[soroban]

Hello, Matt2015!

Determine the general equation of the secant from any point (a,f(a))
to the point on f(x) = 4^x where x = 2.

We are given a point \(\displaystyle (a,f(a)) \,=\,(a,4^a)\)
\(\displaystyle \quad\)and another point \(\displaystyle (2,4^2)\,=\,(2,16).\)

The slope of the secant is: \(\displaystyle \; m \:=\:\dfrac{4^a-16}{a-2}\)

The equation of the secant is: \(\displaystyle \; y - 16 \;=\;\dfrac{4^a-16}{a-2}(x-2) \)

\(\displaystyle \quad y \;=\;\frac{4^a-16}{a-2}\,x - 2\left(\frac{4^a-16}{a-2}\right) + 16\)

Describe how you would use this to find the equation of the tangent at the point (2,f(2)).

Take the limit of the secant equation as \(\displaystyle a\to 2.\)
This will require some Calculus.

 

[Matt2015]

okay, hold on a second. I think this is where I get lost.

I understand fully that the slope of the secant line is formed and ends up being m= (4A-16)/(A-2)

Now I also understand the equation of the line is y=m(x-p)+q which resembles transformations done on functions. Because the equation we are trying to form uses point (2,16) does the equation of the graph not end up being

y = ((4A-16)/(A-2))(x-2)+16 ?? this is the only answer that makes sense to me.

How did you end up with [FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]=[/FONT] ((4A-16)/(A-2)[FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT] ((4A-16)/(A-2)[FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]16[/FONT]

 

[Steven G]

okay, hold on a second. I think this is where I get lost.

I understand fully that the slope of the secant line is formed and ends up being m= (4A-16)/(A-2)

Now I also understand the equation of the line is y=m(x-p)+q which resembles transformations done on functions. Because the equation we are trying to form uses point (2,16) does the equation of the graph not end up being

y = ((4A-16)/(A-2))(x-2)+16 ?? this is the only answer that makes sense to me.

How did you end up with [FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]=[/FONT] ((4A-16)/(A-2)[FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT] ((4A-16)/(A-2)[FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]16[/FONT]

m(x-2) is NOT mx-2!. Rather m(x-2)=mx-2x and you know what m is.

 

[Matt2015]

I still dont understand it. All of the examples and lessons I have going through my course work simply puts M into the equation of Y=M(X+p)+q Where M = Slope p = x1 and q =y1 which makes it look identical to every other transformation of a function.

A line with slope of M hits the point 2,16 (2 units right , 16 units up) so the equation = y= M(X-2)+16 NOT M(x-2) M+16

 

[Deleted member 4993]

I still dont understand it. All of the examples and lessons I have going through my course work simply puts M into the equation of Y=M(X+p)+q Where M = Slope p = x1 and q =y1 which makes it look identical to every other transformation of a function.

A line with slope of M hits the point 2,16 (2 units right , 16 units up) so the equation = y= M(X-2)+16 NOT M(x-2) M+16

The tutor is NOT saying that! The tutor is saying:

y = M(x-2) + 16 = M*x - M*2 + 16 .... using distributive property of the multiplication.

Take paper and pencil out - write it out. Don't just stare at the screen and make-up mistakes!!