I can't solve this one either! (calculus)?

Question

I only get 1 more chance! Please help.

The equation of the tangent line at (4, f(4))

f(x)= 7x^2

y= [blank]x + [blank]

Answer 1

first find your point
f(4) = 7*16= 112

so you need tangent line through (4,112)

to get the slope find the first derivative of you function and evaluate it for x=4

f' = 14x
f'(4) = 14*4 = 56 = slope

now you have all info you need for a line

y - 112 = 56(x-4)

y = 112 + 56x - 224

y = 56x - 112


double check my numbers!

Answer 2

the eqution of a line is also written
y- y sub 1= m( x- xsub1)

in order to find the equation in the form y= mx+b we need to find x sub1, y sub1 and m

xsub 1 = 4
ysub 1= f(4) = 7(4)^2= 112

the next thing we need to know is the slope or in other words the first derivative
if f(x)=7x^2 then f'(x)=14x
but we need to find the value at x=4 so
f'(4)=56

we go back to our original
y- y sub 1= m( x- xsub1) and substitute
y-112= 56(x-4)
y-112= 56x - 224

move the 112 to the other side and we get

y= 56x-224+112
y=56x-112
or
y= x-2

I hope that was helpful!

Answer 3

the slope of the tangent line comes from the derivative
f'(x)=2*7x
f'(4)=2*7*4=56
the point where the tangent line touches f(x) is
f(4)=7(4)^2=112 --> (4,112)
so now you have the slope of the tangent line and a point on the tangent line. use this formula.
y=mx+b, m=slope, b=intercept
substitute from above.
112=56(4)+b
solve for b= - 112
answer: y = 56x - 112 (i hope!!)

Answer 4

To find the equation of the tangent line, you first need to take the derivative of your function.
f(x)=7x^2
f'(x)=14x
Now put in your x coordinate into the derivative and that gives you the slope at that point
f'(4)=14(4)=56.
Now you have to find the equation using slope intercept form.
y=mx+b where m is the slope and b is the y intercept. y and x come from your coordinates (4,f(4)) which is (4,112)
Now you have
112=56(4)+b
112=224+b
subtract 224 from both sides
b=-112
So your equation is
y=56x-112.

Hope that helps!!

Answer 5

f(x) = 7x^2

We first find the slope of the tangent line at x = 4 by taking the derivative.

f'(x) = 14x

Therefore, plugging in x = 4,

f'(4) = 14(4) = 56

Therefore, m = 56

Note that f(4) = 112, so we want to find the equation of the line through (4, 112).

We use the slope formula to obtain the equation of the line.

(y2 - y1) / (x2 - x1) = m

For (x1, y1), we plug in (4, 112).
For (x2, y2), we plug in (x, y)
For m, we plug in our calculated value, 56.

(y - 112) / (x - 4) = 56

Multiplying both sides by (x - 4),

y - 112 = 56(x - 4)

And now, putting this in y = mx + b form,

y - 112 = 56x - 224
y = 56x - 112

Answer 6

f(4) = 7(16) = 112
Eqn of needed line, y(x) = ax + b, where a,b are constants
f'(x) = 7(2x) = 14x
f'(4) = 14(4) = 56 the slope or a in y(x) above
y(x) = 56x + b
y(4) = 56(4) + b = f(4) = 112
b = 112 - 56(4) = -112

Answer 7

gradient of the tangent at point (x,f(x)) is given by the first order derivative of the function.
so first differenciate the function and determine
f'(4) by substituting x=4
now u know the gradient and a point ( 4,7*4^2) on the tangent line
so u can find the line now
go ahead best of luck

Answer 8

Actually, use the point-slope formula instead, where the point is (4, f(4)) and the slope is f'(4), and convert to slope-intercept form.

Answer 9

slope=dy/dx=14x at x=4 slope =56

equation y-f(4)=slope(x-4)