how do I show that y=11x+5 is a tangent to the curve y=3x squared+5x+8?

Answer 1

we want to show that y=11x+5 is a
tangent to the curve y=3x^2+5x+8

first determine the point(s) where the
line y=11x+5 cuts the curve
y=3x^2+5x+8
y=y,therefore,
11x+5=3x^2+5x+8
3x^2-6x+3=0
divide by 3
x^2-2x+1=0
(x-1)(x-1)=0
>>>x=1
hence the two curves meet
at the point where x=1
substitute x=1 into any of the
two equations in order to obtain
a corresponding value of y
y=11x+5 >>>>>y=16

we now know that the two
curves meet at the point
(1,16)-we still do not know
if that is a tangent to
y=3x^2+5x+8

the tangent is a line of gradient
(dy/dx)1 passing through the
point (x1,y1)-it's standard equation
is:-
y-y1=(dy/dx)1{x-x1}.....(1)
where x1 is substituted for x
and(dy/dx)1 is the value
of dy/dx when x1 is substituted
let point(x1,y1)=(1,16)
d(3x^2+5x+8)/dx=6x+5
giving (dy/dx)1=11
now substitute these values
into (1)
y-16=11(x-1)
y=11x-11+16
y=11x+5

therefore,y=11x+5 is a tangent to
the curve y=3x^2+5x+8 QED

if in doubt of this answer,i will
explain it further-thanks for the
question

Answer 2

A lot of the materila you have been given in the answers is correct but your question does require you to show that
y = 11x + 5 is a tangent to the curve. It is not sufficient to caculate the coordinates of the point where the line touches the curve because this assumes that the line is a tangent and that is what you have been asked to show.

Your approach should be to demonstrate this by accurately drawing the parabolic curve y = 3x^2 +5x + 8 and then the straight line y = 11x + 5. The line will touch the curve at a single point (a tangent) and from the graph you can ascertain the coordinates of this point. This can be checked by algebra and calculus as follows:

The curve y = 3x^2 + 5x + 8 is a parabola. Tangents to this curve only touch the curve at a single point. The equation of these tangents is found by differentiating the quadratic equation.

y = 3x^2 + 5x + 8 so dy/dx = 6x + 5

The equation y = 11x + 5 is the equation of a straight line and the gradient is 11.

Hence for this specific tangent:

6x + 5 = 11

6x = 11 - 5 = 6

x = 6/6 = 1

When x = 1, y = 11*1 + 5 = 16

So the coordinates of the point where the straight line y = 11x +5 touches the parabolic curve y = 3x^2 +5x + 8 are (1,16).

Answer 3

Now, to show that a line is a tangent to a particular curve, we need to show that the value of the derivative at the point of tangency of the curve is equal to the slope of the line.

So, in plain English, y = 3x^2+5x+8

Its derivative at a generalized x,y will be, dy/dx=6x+5.

So, if this derivative has to be equal to the slope of the line given as 11x+5, then 6x+5=11, or x=1.
the y value is 16 (obtained by putting x=1 in the line equation).
This (x,y) must lie on the curve, so we check that by computing 3.1^2+5.1+8 = 16 = y given. So, the point of tangency is confirmed.

This back check is critical because for curves of degree >2 (say 3), the derivative equation will be of degree 2, and hence throw up two solutions for x, only ONE of which will be the tangent line.

trust this helps...

Answer 4

You need to show that the curve and the line have the same slope at the same point. However, you do not know what that point is.

First calculate the slope of the curve by taking the derivative.

y = 3x² + 5x + 8
dy/dx = 6x + 5

The slope of the line y = 11x + 5 is obviously 11 everywhere. So calculate where the slope of the curve is also 11.

dy/dx = 6x + 5 = 11
6x = 6
x = 1
y = 3x² + 5x + 8 = 3(1²) + 5(1) + 8 = 3 + 5 + 8 = 16

The curve has slope 11 at the point (1,16).

Now check to see if the line also goes thru that point.

y = 11x + 5 = 11(1) + 5 = 16

So the line also goes thru the point (1,16).

Since both curve and line go thru the point (1,16) and have the same slope there, the line is tangent to the curve at (1,16).

Answer 5

First calculate the slope of the curve by its derivative.

y = 3x² + 5x + 8
dy/dx = 6x + 5

The slope of the line y = 11x + 5 is 11

dy/dx = 6x + 5 = 11
6x = 6
x = 1
Now,
y = 3x² + 5x + 8 = 3(1²) + 5(1) + 8 = 3 + 5 + 8 = 16

The curve has slope 11 at the point (1,16).

Check

y = 11x + 5 = 11(1) + 5 = 16

So the line also goes thru the point (1,16).

Answer 6

Tangent simply means that two lines share one AND ONLY one common point (I am assuming in two dimensions, since you have not specified your level of math).

Therefore, set the two y= equations equal to each other, ie. 3xsquared + 5x + 8 = 11x + 5

You will end up with xsquared + 2x +1 = 0; factor to solve for the point of tangency.

Answer 7

from the line m=11=y'


tangengy point
y=y
discriminant=0

3x^2+5x+8=11x+5

hence x=1

y'=6x+5

for x=1 ,
y'=6*1+5=11
proved.

Answer 8

y=11x + 5
y= 3x squared + 5x + 8

11x+5 = 3x squared + 5x + 8
3x squared + 5x + 8 - 11x - 5 = 0
3x squared - 6x + 3 = 0

3x squared - 3x - 3x + 3 = 0

3x(x-1) -3(x-1) = (3x-3)(x-1)

3x-3=0 , 3x=3, x=1

x-1=0, x=1

Answer 9

is this for an exam paper? Then do what I do. First of all write your name in big flowery writing over the top of the page ... add doodles, make the dots over the 'i' into little flowers. Then draw a rainbow in the corner of the page, a house on the bottom right and nice garden in the left. Spend the remaining exam time colouring alternate lines on the page in rainbow colours. That should fill up the exam time nicely ... hand in the paper. hope the examiner is so impressed with your artwork that he doesn't notice the complete absence of a mathematical answer.
Good luck

Answer 10

the first derivative of the curve should equal the tangent line. apparently in this case it's not

oh, and also the slope of both formulas should be negative reciprocal of each other.