[whatevaxd]

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 :)

[Perito]

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.



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:

Remember that the "y" was solved for above.

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

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).

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)

therefore
therefore

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.

[Unknown008]

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!

[whatevaxd]

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. :(

[Perito]

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.








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

This can be factored to give

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)

because f(q)=g(q)

which can be rearranged and factored to give



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