<p>What is the set of all values for b for which the graphs of y= 2x +b AND y^2=4x intersect in two "distinct" points???</p>
<p>any help will be appreciated, thks</p>
<p>The first equation is a straight line with slope 2 and intercept b. If we are talking about real numbers, the second equation exists only for non-negative values of x and y, and is monotonically increasing for x values starting at 0.</p>
<p>Since y^2 is non-negative, a straight line that intersects the y-axis below the origin will intersect the second graph in only one place. At some point, with a large enough positive b, the two will not intersect at all. Your job is to find the point which we will call P where the line y=2x + b is tangent to the curve y^2 = 4x. Hint: find the x and y coordinates at which the slopes are equal and plug those values into the first equation to solve for b. If we call this value B, then your answer is [0,B).</p>
<p>line y=2x + b is tangent to the curve y^2 = 4x
so, y^2=4x is x=y^2/4 derivative of that is y/2
then what do i do???</p>
<p>You want to put the second equation in the form</p>
<p>y = 2 x^(1/2)</p>
<p>set the derivative of that equal to 2.
i.e. find the value of x where y' = 2.
Plug that value of x into y = 2 x^(1/2) to get the corresponding y value. You now have the coordinates where the line intersects the curve in exactly one spot.
Plug those values of x and y into the equation y=2x + b to get the value of b.</p>
<p>Also note that this is a potential time-waster question on the AP exam. If this was a multiple choice question and only one of the answers has the form [0,B) for a positive value of B, then you could pick that answer without actually having to do the derivative and be on to the next question in well under a minute.</p>
<p>SO IS this right:
y=2x^(1/2)
y'=1 / x^(1/2)
1 / x^(1/2) = 2
so x=1/4
therfore y=1
so (1/4,1) is the coordinate, and i plug those coordinates into the equation y = 2x + b and b=1/2 (Please check this)</p>
<p>Okay so now the question was asking: "What is the set of all values for b for which the graphs of y= 2x +b AND y^2=4x intersect in two "distinct" points???"</p>
<p>how do i go about from here? what is the set of all values of b? 2 distinct points???</p>
<p>The calculation is correct. The set of all values of b is all points from 0 to 0.5, including 0 but not including 0.5. This can be written [0,0.5) or as 0 <= b < 0.5. </p>
<p>Visualize the two graphs, keeping the curve where it is and allowing the line to slide up the y-axis. As long as the y-intercept of the line is less than 0, it intersects the curve in one place, out to the right of the origin. As the line moves up, the intersection point slides closer in toward the origin. When the y-intercept (i.e. b) of the line hits 0, the two graphs now intersect in two places, (0,0) and (1,2). As the line continues upward, the two points approach each other until they meet at (1/4,1) where b=1/2. There is only a single point of intersection there, so the condition is no longer satisfied. As the line continues upward, there is no intersection at all.</p>
<p>but why is it [0,.5) i understand the values of b must be less than .5 if the two curves were to intersect at 2 points. Therefore shouldnt it be [- infinity, .5)<br>
since the curve is y^2=4x and the line is y=2x +b , just graph it and you see that b can be any negative value ex.) y=2x -3 Intersections are (3.33,3.646) and (.677,-1.646) </p>
<p>please let me know if am correct, i appreciate your help thus far</p>
<p>This gets a bit philosophical. Strictly speaking, what you are graphing is not y^2 = 4x but rather the combination of y = +2 |x^(1/2)| and y = -2 |x^(1/2)|. If we are dealing with real rather than imaginary numbers, y^2 will only have positive values and its graph would lie entirely in the first quadrant. What you have graphed contains values in the first and fourth quadrants, with symmetry about the x-axis. This is not actually a function because it contains two values of y for all x > 0. However, depending on how the problem was presented I think you might make an argument for either answer.</p>
<p>i better ask some teachers tommorow, i let you know what happens, thanks for your help thus far</p>
<p>I suggest you go with whatever the teachers say. I probably haven't taken a math test in your lifetime.</p>
<p>my calc teacher said that the values of b could be negative since the questions gives the function y^2 = 4x, the question doesnt state whether or not the the equations must be functions, therefore answer is [-infinity,1/2) </p>
<p>thanks for your help in solving this problem :), regardless of whether or not you have taken a math test in my lifetime, you are one clever man</p>
<p>i hope u dont mind me asking 1 more question, this one is rather more straightforward</p>
<p>Given: f(x)= absolute value of sinx for which x is greater than or equal to negative pi and less than or equal to pi</p>
<p>and g(x)= x^2 for all real x</p>
<p>what is the domain and range of H for which H(x)=g(f(x)))</p>
<p>so, H(x) = (absolute value of sinx)^2</p>
<p>my Domain was : x is greater than or equal to negative infiinity and less than or equal to infinity</p>
<p>my Range was y is greater than or equal to 0 and less than or equal to 1</p>
<p>isn't my Domain and Range correct? apparently my Domain is supposedly wrong but how is that possible? what have i done wrong? any help will be appreciated</p>
<p>I know the range is right, and i'm pretty sure that the domain is also correct, but what do i know.</p>
<p>yea i thought so the range was correct but the domain was marked wrong, i dont know why and my teacher wont tell</p>
<p>Think of domain as the input of a function and range as its output. In this case, you are taking the output of function f and using it as the input of function g. Therefore, the range of function f is the domain of function g.</p>
<p>In this case, f(x)=abs(sin(x)) with a specified domain of -pi to +pi.
The range of f(x) is [0,1]. That is also the domain of g(x).
Since g(x) = x^2, the range of g(x) is also [0,1].</p>
<p>so what would be the domain and range of H(x)=g(f(x)))???
Range [0,1]
Domain[-pi,pi] </p>
<p>is that right?</p>
<p>Now you've got it. The domain is that of f(x) and the range is that of g(x). Note that had g(x) been undefined anywhere, you would have had to exclude that part of the domain. That obviously does not apply here.</p>
<p>since you have been so helpful im going to ask you one last final question, i promise no more questions</p>
<p>i need to find the value of m for when x'=0 (note that m is positive)</p>
<p>the equation is y = mx - [(e^(2m))(x^2) / 1000 ]</p>
<p>i dont know how to solve for x' because there are two x's? and when i factor 1 x out i end up with 0=0. got any ideas???</p>
<p>OK, now you've gone and confused me. Normally in this kind of problem, m is a constant, so it would have the same value for all x and y and the "when x'=0" part would be irrelevent. Are you certain that you are not being asked to find x in terms of m when y'=0? That would be pretty straightforward and the type of problem that is very common in first year calculus.</p>
<p>If m is indeed a variable, then does x' mean dx/dy or dx/dm? Is there more to the statement of the problem than you have given here?</p>