Differential Calculus?

Question

I have a summative test tomorrow and I can't seem to understand HOW to do this question (I know the answer):

Find the equation of the tangent to y = x^2 - x + 9 at the point where x = a.
Hence, find the equations of the two tangents from (0,0) to the curve. State the coordinates of the points of contact.


The answer is y = (2a - 1)x - a^2 + 9.
y = 5x, contact at (3, 15)
y= - 7x, contact at (-3, 21)

Thankyou so, so much to whoever can show me how to work out the answer!

Answer 1

The slope at x=a is just the derivative of the equation:

dy/dx = 2x - 1
m(a) = 2a - 1

The y-value of the curve where x=a, is just the equation with "a" replacing "x": a^2 - a + 9.

The equation for a line with slope 2a-1, that passes through that point (a, a^2 - a + 9) is:

y = mx + b
y = (2a-1)x + b

(a^2 - a + 9) = (2a-1)a + b
a^2 - a + 9 = 2a^2 - a + b
b = (9 - a^2)

Since we now know b, we can substitute back into the slope-intercept form:

y = mx + b
y = (2a -1)x + (9 - a^2)
y = (2a - 1)x - a^2 + 9

So now you have the equation for the tangent to the curve where x=a. You want to know the tangents that pass through (0,0), i.e., where the Y-intercept is 0.

solving for "b = 0", we find the a-values that qualify:

9 - a^2 = 0
a^2 = 9

a = 3, a = -3

So, the two tangents that pass through (0,0) are when a=3 and when a=-3. Remembering that a is the x-value of the point at which the tangent touches the curve, the equations are:

(1) when a=3
y = mx + b
y = (2a - 1)x + 0
y = 5x

(2) when a=-3
y = mx + b
y = (2a - 1)x + 0
y = -7x

Now, to compute the Y-value of those data points, use the original equation y = x^2 - x + 9:

(1) when x=3, y = 3^2 - 3 + 9, which is 15

(2) when x=-3, y = 3^2 - (-3) + 9, which is 21

So...

(1) the tangent that contacts at x=3 has equation y = 5x and contacts at (3,15)

(2) the tangent that contacts at x=-3 has equation y=-7x and contacts at (-3,21)

Answer 2

the tangent equation may be something like this y(tang)=mx+c, where m=y'(a)
y'=2x-1 hence m=2a-1 and y(tang)=(2a-1)x+c
Let's find c by y(x=a)=y(tang x=a)....c=9-a^2.
And now only replace c in y(tang) then:
y(tang)=(2a-1)x+9-a^2.

To find the 2 tangents from (0,0) only have to think that c=0 son it let you find a.
9-a^2=0, a= + 3 and a= - 3. Replace these values in y(tang) and finally,

y1=(2·3-1)x=5x
y2=(2·(-3)-1)x=-7x

By knowing the x coordinate (x=a) you can find the y coordinate replacing 'a' values in the original equation:

for y1, a=3: y(x= + 3)= 3^2-3+9=15 so the point is (3,15)
for y2, a=-3: Y(x=-3)= (-3)^2-(-3)+9=21 so the point is (3,21)

Hope this will help!
Greetings

Answer 3

I was going to say that it is possible you could pass if you were exceedingly drone like as a student and took everything everybody said without question and as fact, but really that does not even help you. Anyone would be beyond confused in this course if they do not understand what a derivative is, let alone an integral. Some of the course is vectors, but the rest is a challenge of if you can really understand the calculus enough to write down and figure out which integral is the relevant one necessary for particular applications. Can you do that? Probably not if you do not even understand integrals. Even students that have taken calc 1 and 2 barely grasp the idea, calc 3 can be a very difficult class if your university is hard enough. It just is not going to work out though for most anyone. Figure out your schedule and make some sacrifices, failing a class would be silly this early on.

Answer 4

y=x^2-x+9
slope of tangent =dy = 2x -1
the slope at x = a is 2a-1
when x = a y =a^2-a+9
the tangent must pass through this point. use point slope form
y-y1=m(x-x1)
y-a^2+a-9 =(2a-1)(x-a)
y =2ax -x -2a^2 +a +a^2-a +9 now collect terms
y =(2a-1)x -a^2 +9

on the last two problems you know two points so you can calculate the slope then use the second point in the point slope equation:
m=(y2-y1)/(x2-x1)=(15-0)/(3-0) =5
y-15 =5(x-3)
y=5x-15 +15
y=5x

m=(21-0)/(-3-0) = -7
y-21=-7(x+3)
y-21 =-7x -21
y= 7x

Answer 5

something is not written correctly
check the question again and repost
=)