Question 73103: having problems/please help
1.find the x-intercept and y-intercept of the equation 5x-3y=30.
2.find an equation for the line slope 1/2 and y-intercept 3.
3.find an equation for the line with y-intercept 3 that is perpendicular to the line y=2/(3)x-4.
4.if f(x)=x^2 + 5,find f(a+h)-f(a).
5.If P(4,-5) is a point on the graph of function y= f(x),find the corresponding point on the graph of y=2f(x-6).
6.find the maximum value of y=x^2 +6x.
Answer by checkley75(3666)   (Show Source):
You can put this solution on YOUR website! 1) 5X-3Y=30
-3Y=-5X+30
Y=-5X/-3+30/-3
Y=5X/3-10 THEREFORE THE Y INTERCEPT IS -10 OR (0,-10) & HAS A SLOPE OF 5/3.
TO FIND THE X INTERCEPT WE NEED TO DIVIDR 10BY 5/3 THUS
10/(5/3)
10*3/5
30/5
6 IS THE X INTERCEPT OR (6,0)
--------------------------------------------------------
2) SLOPE OF 1/2 & Y INTERCEPT OF 3
Y=X/2+3
----------------------------------------------------------
3) Y=-3X/2+3 THIS PERPENDICULAR LINE HAS A SLOPE OF THE NEGATIVE RECIPRICAL OF THE LINE IT IS PERPENDICULAR TO. OR THIS SLOPE IS -3/2
|
|
|
