Ask Me Help Desk - Linear relationships

1.Find the coordinates of the point where the line 2y=x-4 cuts the x axis.
Find the equation of the straight line through this point and (-1,10)

Are the coordinates and the equation of this question connected? I have done this exercise few months ago but I have forgotten everything. Some explaining would be nice.. :)

2.Find the coordinates of the point where the line 2y=-x+6 cuts the x axis.
Find the equation of the straight line through this point and the point (8,8)

3.If f(x)=2x+10 and g(x)=x^2-3x-4
Find
a.f(2)+g(2)

b.f(x)+g(x)

c.f(2x)+g(2x)

d. the values of p for which g(p)=0

e.The values of q for which f(q)=g(q)

Thx in advance :)

Quote:

1.Find the coordinates of the point where the line 2y=x-4 cuts the x axis.
Find the equation of the straight line through this point and (-1,10)

Are the coordinates and the equation of this question connected? I have done this exercise few months ago but i have forgotten everything. Some explaining would be nice..

The point where a line cuts the x-axis is the point where x = 0 (because x = 0 at the x-axis). All you need to do is replace "x" in your equation with "0" and solve for y.

Taking the point (0,y) you solved for, create an equation for the line.

$slope = \frac {\delta Y}{\delta X} = \frac {10-y}{-1-0} = y-10$

The standard slope-intercept form of a line is "Y=mX + b" where "m" is the slope and "b" is the y-intercept. Plug in the slope into the line:

$Y=(y-10)X + b$ Remember that the "y" was solved for above.

Enter one point, say (-1,10), for X and Y and solve for b.

Quote:

2.Find the coordinates of the point where the line 2y=-x+6 cuts the x axis.
Find the equation of the straight line through this point and the point (8,8)

This is done exactly the same way as #1. The value of "x" is 0 at the point where a line cuts the x-axis (the x-intercept).

Quote:

3.If f(x)=2x+10 and g(x)=x^2-3x-4, Find

a.f(2)+g(2)
b.f(x)+g(x)
c.f(2x)+g(2x)
d. the values of p for which g(p)=0
e.The values of q for which f(q)=g(q)

$f(x) = 2x+10$ therefore $f(2) = 2 \times 2 +10 = 14$
$g(x)=x^2-3x-4$ therefore $g(2) = 2^2-3 \cdot 2-4=-6$

The others are straight-forward. Give them a try. If you have trouble, post back with what you've done and I'll help you more.

1. You firstly have to find the point of intersection of your line with the x axis Now, find the gradient of the line that passes through the point you just found, and the given point. With the gradient and at least a point on the line, you can easily find the equation of that line.

2. Same as for 1.

3.

a. Find f(2) by replacing x by 2 in f(x) and g(2) by replacing x by two in the given g(x). Then, add both values.

b. Just add these two functions. (2x+10)+(x^2-3x-4)

c. Same as for a. but putting 2x instead of 2.

d. Replace x by p in g(x), and set that expression to zero. Solve for p by factorising.

e. Replace q in f(x) and g(x) and equate the two expressions.

Hope it helped! :) If you still have difficulties, don't hesitate to post a reply!

for question 3c.
I managed to get
4+20+2x^2-6-8=2x^2-2x+6
but the answers at the back of the book is 4x^2-2x+6
can u explain how do get 4x^2?

I still don't get how to get 3d and 3e. :(

Quote:

for question 3c, I managed to get

4+20+2x^2-6-8=2x^2-2x+6

but the answers at the back of the book is 4x^2-2x+6
can u explain how do get 4x^2?

I still don't understand 3d and 3e.

3c.
$f(x)=2x+10\,and\, g(x)=x^2-3x-4$

$f(2x) = 2(2x) + 10=4x+10$

$g(2x) = (2x)^2-3(2x)-4=4x^2-6x-4$

$f(2x)+g(2x)=(4x+10)+(4x^2-6x-4)=4x^2-2x+6$

3d. The values of p for which g(p)=0

$g(p)=p^2-3p-4=0$ This can be factored to give

$(p-4)(p+1)=0$ and the roots of this equation are p=4 and p=-1. These are the values that are requested.

3e.The values of q for which f(q)=g(q)

$f(q)=2q+10 = g(q)=q^2-3q-4$ because f(q)=g(q)

$2q+10 =q^2-3q-4$ which can be rearranged and factored to give

$q^2-5q-14=(q-7)(q+2)=0$

The roots of the equation are q=7 and q=-2. Therefore, these are the values of q for which f(q) = g(q).