2017年8月3日 · A straight line parallel to the x-axis through the point (0,1) y=1 A straight line in slope (m) and intercept (c) form is: y=mx+c In this example, y is a straight line of slope 0 and y-intercept of 1 :. y = 0*x + 1 Hence, the graph of y a is straight line parallel to the x …

2017年1月29日 · How do you graph #y=-1#? Algebra Graphs of Linear Equations and Functions Graphs Using Slope-Intercept Form.

2017年12月18日 · Course: Differential Equations with Linear Algebra (don't use Linear Algebra please)

You can put this solution on YOUR website! What the question is asking you to do is to solve for x. This means that you will need to get x on one side of the equation and everything else on the other side.

Now mark these two locations on your graph paper. Starting at the origin of your graph (where the x-axis crosses the y-axis), go to the right 2 squares (x=2) then up 1 square (y=1) and mark your first point. For the second point, again, start at the origin and go right 5 squares (x=5) and then up 4 squares (y=4) and mark your second point.

You can put this solution on YOUR website! y > x - 1 Graph the boundary line y = x - 1 Slope-intercept form is y = mx + b

You can put this solution on YOUR website! y-2 > 1 y-2+2 > 1+2 y+0 > 3 y > 3 To graph y > 3, you step 1) draw out a number line

2017年1月17日 · see below Left Hand Side: Use identity :1+tan^2y=sec^2y cot^2y(sec^2y-1)=cot^2ytan^2y =cos y/siny * siny /cos y =cancelcos y/cancelsiny * cancelsiny /cancelcos y =1 :.= Right Hand Side Trigonometry Science

Let's take the function #y=1/x# graph{1/x [-10, 10, -5, 5]} You will see, that the larger we make #x# the closer #y# will be to #0# but it will never be #0# #(x->oo)# In this case we call the line #y=0# (the x-axis) an asymptote. On the other hand, #x# cannot be #0# (you can't divide by #0#) So the line #x=0# (the y-axis) is another asymptote.

2015年12月6日 · f^(-1)(x) = ln x +1 Given y= e^(x-1) (this is a one-to-one function) Step 1: Switch x for y and y for x like this color(red) x= e^(color(blue)(y)-1) Step 2: Begin to solve for y Take ln of both side ln(x) = lne^(y-1) Use the properties of log ln e= 1 ; log a^n = nlog a ln x= y-1 ln x + 1 = y Step 3: Change y to f^(-1)(x) f^(-1)(x) = ln x +1 *Note: To solve exponential equation, we use ...