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Focus (0,1) and directryx is y=x-2 , graph and give equation

Please show how to solve and graph this type of problem.

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Finding the Vertex, Focus, and Directrix of a Parabola Given ...

Home / Blog / Finding the Vertex, Focus, ... the directrix equation is y = 0 – (-1) = 1. The focus is at ... questions about its graph, such as the parabola’s ...
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Identify the focus, directrix, and axis of symmetry of x=2y^2. Graph the equation.

This is the equation of a parabola, but it's lying on its side arms on either side of the x axis, which is the axis of symmetry, and the arms are on the positive side of x. The parabola's vertex is at the origin. The focus lies along the axis of symmetry and the directrix is a line x=-h. The rule about a parabola is that any point on the curve is equidistant from the directrix line and the focus point. Because of this rule when y=0, x=0 so the origin is midway between the directrix and focus, therefore the focus is at (h,0). Take any point P(x,y) on the curve. It's distance from the directrix is h+x; its distance from the focus is given by Pythagoras: sqrt((h-x)^2+y^2). But we know the equation of the parabola is x=y^2 so we can replace x and: h+2y^2=sqrt((h-2y^2)^2+y^2). Squaring both sides: (h+2y^2)^2=(h-2y^2)^2+y^2, y^2=(h+2y^2-h+2y^2)(h+2y^2+h-2y^2) (difference of two squares), y^2=8y^2h, h=1/8, because the y^2 cancel out. So the focus is at (1/8,0) and the directrix line is x=-1/8. You've got the general picture of how the parabola looks, so you can mark the focus and directrix line, you know the vertex is at (0,0), and you can plot a few points to help you get the curve right. When y=1 and -1 x=2 and when y=2 and -2 x=8. That gives you 4 points: (2,1), (2,-1), (8,2), (8,-2). See how the x axis acts like a mirror reflecting each half of the curve.
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sketch the graph of x^2+2x+4y-15=0. give the coordinate of the focus, vertex ends of latus rectum and equation of the directrix

x^2+2x+4y-15=0 is the same as x^2+2x+1-16+4y=0. So (x+1)^2=16-4y and 4(y-4)=-(x+1)^2. This is in standard parabola form (4f(y-k)=(x-h)^2 where (h,k) is the vertex and |f| is the focal length) with vertex at (-1,4). The parabola is inverted with axis x=-1 on which lies the focus, which is the same distance from the vertex as the directrix. The focus is on the inside of the curve while the directrix line is outside the curve and perpendicular to the parabola's axis. The latus rectum (the chord of the parabola) is parallel to the directrix and passes through the focus. The focal length in this case is 1 and because it lies within the inverted curve its coord of the focus is (-1,3). The directrix line is above the curve at y=5. The latus rectum is at y=3 with endpoints where this line intersects the curve: -4=-(x+1)^2, x+1=±2 so the endpoints have x-coords ±2-1=1 and -3: (1,3) and (-3,3). The parabola is the red curve and the blue line is the directrix line. The green line spanning the parabola is the chord (latus rectum) but is shown extended beyond the curve.
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Focus (0,1) and directryx is y=x-2 , graph and give equation

wotz that "fokus" & "direktree" stuff????? y=x-2 is a strate line with slope=1  (angel=45 deg) Maebee yu want it tu go thru point=(0,1)????????? if so, y-intersept=1, so line formula is y=x+1
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how do we graph this equation 3x+4y=20

how do we graph this equation 3x+4y=20 3x+4y=20 need to graph it with three points. Solve the equation for y. It will then be in slope/intercept form. 3x + 4y = 20 4y = -3x + 20 y = -3/4 x + 5 That gives us one point, the y-intercept: (0, 5) Now, choose two values for x, plug them into the equation, and solve for y. Use, x = 20. y = -3/4 x + 5 y = -3/4(20) + 5 y = -15 + 5 y = -10 That gives us the point (20, -10) Now use x = -20, to keep it simple. y = -3/4 x + 5 y = -3/4(-20) + 5 y = 15 + 5 y = 20 That gives us the point (-20, 20) Plot those three points on the graph and draw a line through them.
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word problem please show how

Suppose that in one week, your company takes in $2,220 in sales at one of its cell phone stores. The price breakdown per phone is as follows: Cell phone model A4: $35 Smartphone Z20: $50 Suppose that a total of 51 phones were sold. Set up the system of equations that needs to be solved to determine how many of each type of phone were sold. Give a clear definition of the variables in the system. Solve the system of equations, showing clearly how the solution was determined, and state the results clearly in light of the real-world situation. Verify your results of the 2 linear equations by graphing in the desired graphing program and paste the graph in your assignment document (edit/copy image). Explain how the results are verified by the graph.
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Give the slope-intercept form of the equation of the line that is perpendicular to and contains (-1, 9)

Start by solving for y to get this in y=mx+b form subtract 5x from both sides 4y=-5x+4 divide by 4 y=4/5x+4/4 simplify to y=4/5x+1 Now you know that the slope of the first line is 4/5 The slope of the second line is -4/5 so the second equation is y=-4/5x+? since the line has to cross the point -1,9 you can plot that point on the graph and then use the slope to find another point. from the point -1,9 you can use rise over run to go down the graph 4 spots to 5 and to the right on the graph 5 spots to 4 your second point would be (4,5) y=-4/5x+41/5 is the equation for the second line connect those points to get your line.  The two lines look like this:
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how to find slope and y-intercept of y=3x-5

Any line equation which is in the form y=m*x + b is called slope-intercept form. What that gives you is the slope is the number which is multiplied by X (called the coefficient) while the slope intercept is the Y value when X=0. So if we take your first equation: y = 3x - 5 The slope (or m) = 3 The Y intercept = -5 Slope is defined as either "the change in y divided by the change in x" or "rise over run", so that 3, really can be considered as 3/1. Each change in X of 1 will change Y by 3. So slope = 3/1. Graphing any line can come from 2 methods. 1) create a table of values or 2) calculate a single point (x,y) and then apply the slope to find a second coordinate pair (x,y). For the equation y=3x - 5: y = 3x - 5 x y 0 -5 1 -2 Replacing the values for X into the original equation, will come out to the values for Y. So when X=0, y = 3*0 - 5, or simply -5.  When X = 1, Y = 3*1 -5 or simply -2. With these two coordinate pairs of points, you can plot a dot on your graph at each (0,-5) and (1, -2) then draw a straight line which goes through each point and continues straight in each direction, probably ending each end of this line with an arrow to show it continues. I do not have a way to include a picture here of a graph. The second way to graph a line is as follows. You need a starting point that will be on the line. Given the form of y=mx + b, you have a simple point which can be used at the y-intercept. The point is always in the form of (0, b), so in this case it is (0,-5). From that first point, you will apply your slope. The slope is 3 (or technically 3/1 which is a big help). From the initial point (0,-5) you will go UP 3 and RIGHT 1 and that will be the next point that is easy to find. Connect those two points and continue the line in each direction and that will be a graph of your line. Anytime your slope is positive, you will use it by going UP the top number (numerator of the slope) and going RIGHT the bottom number (denominator of the slope). But if your slope is negative (like your second problem is) you will use it by going DOWN the numerator and then RIGHT the denominator. The equation is y= -2/3x + 4 ( / = divided). I need to state the slope and y-intercept. I will not walk through the details on the second equation, but you should have enough information to get the answer from the above example.
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Find the vertex, focus, and directrix of the conic section 2x^2+12x+16y+2=0

The equation can be divided through by 2: x^2+6x+8y+1=0; x^2+6x+9-9+1+8y=0; (x+3)^2-8+8y=0; -8(y-1)=(x+3)^2. The vertex is at (-3,1) and the curve is a parabola like an inverted U. The vertex line is x=-3, and the focus lies on this line. The properties of a parabola make the following true: (y-d)^2=(x-h)^2+(y-k)^2, where (h,k) is the focus and y=d is the directrix. The rule is that all points on the parabola are equidistant from the focus and directrix line (shortest distance is a perpendicular). So, (y-d)^2-(y-k)^2=(x-h)^2; (y-d-y+k)(y-d+y-k)=(x-h)^2; (k-d)(2y-k-d)=(x-h)^2; 2y(k-d)-(k-d)(k+d)=(x-h)^2. comparing terms we see that (x+3)^2=(x-h)^2, so h=-3; 2y(k-d)=-8y, so k-d=-4; -(k-d)*(k+d)=8, so 4(k+d)=8; k+d=2. Add together the equations containing k and d: 2k=-2, k=-1 and d=3. So the focus is (h,k)=(-3,-1) and the directrix is y=3. Check: (y-3)^2=(x+3)^2+(y+1)^2; (y-3)^2-(y+1)^2=(x+3)^2; -4(2y-2)=-8(y-1)=(x+3)^2 as above. Seems all OK. To draw the graph mark the vertex of the upside down U. That's the highest point. Also draw the vertex line x=-3, which is the axis of symmetry. When y=0 (x axis) x=-3+2sqrt(2)=0.17 approx and x=-3-2sqrt(2)=-5.83, so that's where the curve intersects the x axis. When x=0 y=-0.125. You can also draw the directrix line at y=3 which lies above the curve and the focus which is on the inside of the U.
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A function f(x) is graphed below. Find f(3) and f^-1 (3). f(3)=? and f^-1(3)=?

f(3)=2 and f^-1(3)=2.25 Reasoning: the line has a y coordinate of 2 for the x coordinate of 3. That gives us f(3)=2. The inverse function requires us to find the x coordinate when the y coordinate is 3. This gives us a value a bit bigger than 2. But we can find the equation of the graph. The slope is -6/4.5=-4/3, so the equation is of the form y=-4x/3+c where c is the y intercept, which is 6, so y=-4x/3+6. Put y=3 and we get 4x/3=3 so x=9/4=2.25. That gives us f^-1(3)=2.25.
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how do i graph y=2.5x

The graph is linear (a straight line) because there are no powers of x or y (other than power of 1). You only need two points so that you can connect them by a line. Put x=0 into the equation and you get y=0. That gives you the point (0,0) or the origin where the axes meet. Now put x=4 and y becomes 2.5*4=10, to give you the point (4,10) (the graph is a grid, 4 divisions (columns) to the right and 10 divisions (rows) straight up). Draw a line through (0,0) and (4,10) so it doesn't just join the points, but continues past them, and that's your graph. 
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