Guide :

find the axis of symetryof the graph of the parabola

F(x)=(x-4)^2 a x=0 b x=-4 c x=4 d y=4

Research, Knowledge and Information :


How to Find an Axis of Symmetry: 11 Steps (with Pictures)


How to Find an Axis of Symmetry. The graph of a polynomial or function reveals many characteristics that would not be clear without a visual representation. One of ...
Read More At : www.wikihow.com...

Parabola Intercepts. How to find the x intercept and y intercept


Graph Paper Maker; Number Line ... the point at which the parabola intercepts the y-axis. ... point at which the parabola intersects the x-axis. A parabola can have ...
Read More At : www.mathwarehouse.com...

Graphing Parabolas in Vertex Form - Mesa Community College


Here are the steps required for Graphing Parabolas in the Form y = a(x – h) 2 + k: Step 1: ... Graph the parabola using the points found in steps 1 – 3.
Read More At : mesacc.edu...

Parabolas in Standard, Intercept, and Vertex Form - Video ...


This means you can find it on your graph by working your way ... we can easily tell where the axis of symmetry is ... Parabolas in Standard, Intercept, and Vertex ...
Read More At : study.com...

Three Ways to Find the Vertex of a Parabola - Math Forum


For H(x) = -x2 - 8x - 15, what are the coordinates of the vertex, what is the equation of the axis of symmetry, ... Three Ways to Find the Vertex of a Parabola
Read More At : mathforum.org...

Worked example: intercepts from an equation (video) | Khan ...


Worked example: intercepts from ... We're told to find the x- and y-intercepts for the graph ... the x-intercept is the point on the graph that intersects the x-axis.
Read More At : www.khanacademy.org...

Lines of Symmetry of Plane Shapes - Math is Fun


Folding Test. You can find if a shape has a Line of Symmetry by folding it.
Read More At : www.mathsisfun.com...

SOLUTION: i need to find the x-intercept, y-intercept, axis ...


find the x-intercept, y-intercept, axis of ... The vertex of the general parabola is at the ... The y-intercept is where the graph crosses the y-axis, that is ...
Read More At : www.algebra.com...

SOLUTION: find the vertex, line of symmetry, minimum or ...


negative sign for lead coefficient means parabola opens downward, has a maximum of 28 axis of symmetry: x=5 see graph below as a visual check on answers: .. ...
Read More At : www.algebra.com...

Suggested Questions And Answer :


how to solve area under the curve x = y^2 - 2y , y=0 , x = -1

It's best to sketch the curve first. This helps to picture how to find the area under the curve. It's a sideways parabola. In a way, you can swap the x and y variables so that the graph looks like y=x^2-2x but the axes are interchanged to produce x=y^2-2y. When x=0, y^2-2y=0=y(y-2) so y=0 (origin) and 2 are the y intercepts. When y=0, x=0 (origin). The vertex is when 2y-2 (the derivative with respect to y) is zero, so y=1 and x=-1. The vertex is at (-1,1). The second derivative is 2, so the turning point is a minimum. The graph is U-shaped lying on its side with arms pointing to the right (positive). The parabola has its vertex in quadrant 2 and its arms cross the y axis at 0 and 2. x=y^2-2y, but what is y in terms of x? completing the square we have y^2-2y+1=1+x, (y-1)^2=1+x and y=1+sqrt(1+x). There are generally two values of y for the same value of x. The graph illustrates this, because the parabola is lying on its side so the x value picks up two values of y except at the vertex. We need the lower part of the graph only, given by y=1-sqrt(1+x). If on the graph we paint the inside of the parabola blue and the outside red, this will help us to identify the areas. We want, I think, the red area between the vertex and the x axis and, I assume, up to the y axis, because the curve dips below the x axis when x is positive. Imagine thin rectangles width dx lying vertically with a length equal to y=1-sqrt(1+x). The length of the rectangles for y=1+sqrt(x) include blue and red areas. The area of each rectangle is ydx and the area under the graph is the sum total of the areas of the thin rectangles. When the rectangles are infinitely thin this sum total is the integral of ydx. If we wanted, for example, just the blue area we would need to subtract the red area from the blue-and-red area. This is why it's important to sketch a graph and understand how to apply the mathematics. So our thin rectangles start at x=-1, when y=1, so this rectangle has a width dx and a length of 1; and ends with a rectangle of zero height when x=0. If we integrate between the x limits of -1 and 0 we will get the red area we need: integral((1-sqrt(1+x))dx)=integral((1-(1+x)^1/2)dx)=[x-2(1+x)^(3/2)/3](-1 Read More: ...

y=x^2-4 (find x and y intercepts, vertex and axis) I have the answer, but do not know how to work the problem

You can do this a few different ways. You can graph it on a graphing calculator and use the trace function to find the intercepts.  If you don't have a calculator you can plug in points for y and then solve for x.  You know that this is a parabola because of the x^2.  There are going to be 2 x values because you will have to take the square root. Let's take the example of setting y=0 which will also give us the x intercepts Plug that in and solve for x 0=x^2-4 add 4 to both sides 4=x^2 take the square root of both sides x=+2 and -2 so you have two points 0,2 and 0,-2 Now to find the y intercept you plug in 0 for x and solve for y y=0^2-4 y=-4 you now have a 3rd point -4,0 which is the y intercept.  You can draw the graph with that and see that the parabola is split by the y axis with a vertex at -4,0. Here is the graph.          
Read More: ...

area under y=3(x-2)^2+1

The first step is to sketch the curve on a graph. This curve is a parabola, a U-shaped curve in which the arms spread out gradually. The key points of a parabola are intercepts, one at least, and the vertex, which is where the curve changes direction. The vertical line through the vertex is a line of symmetry acting like a mirror reflecting the two halves of the curve. If we rewrite the equation: y-1=3(x-2)^2 we can find the origin of the parabola. When y=1, x=2 which tells us the origin is (2,1), which is the vertex and the line x=2 is the line of symmetry. When x=0, y=13, so the curve cuts the y axis at (0,13). When y=0 (the x axis) the curve has no real value so it doesn't touch or cut the x axis. This is enough to sketch the curve and actually see the area you need to find.  The next step is to enclose the area by establishing limits. The normal limits will trap an area between the curve and one or both axes. These will become the integration limits so that a definite integration can take place.  Now we build the integral by dividing the area into thin rectangular strips. The length of a vertical rectangle is y and its width is dx. The area is ydx and the sum of the areas of the rectangles gives the area under the curve so we have integral(ydx)=integral((3(x-2)^2-1)dx) and the limits for integration will be a to b, a Read More: ...

graph the quadratic function f(x)=-2x^2-x-2

This is a parabola, an inverted U-shaped curve. It doesn't intersect the x axis because there are no real zeroes. But when x=0, f(x)=-2 so the vertical axis is intercepted at -2. The graph can be written in the form f(x)=a(x-h)^2+k, where (h,k) is the vertex or origin. To find h and k we expand this: f(x)=ax^2-2axh+ah^2+k. From this a=-2, -2ah=-1, so h=-1/4 and ah^2+k=-2, so k=-2+1/8=-15/8 and f(x)=-2(x+1/4)^2-15/8. The vertical line of symmetry is x=-1/4 and the vertex (maximum) lies on this line at (-1/4,-15/8), below the x axis. These are properties of the parabola that help you to draw it. It helps to plot a few other points, for example, (1,-5) and (-1,-3) to get some idea of the spread of the arms of the inverted U. Two other properties of the parabola may also help to graph it: directrix and focus. The directrix is a line defined by a, so that the line is outside the parabola at f(x)=1/(4a)=-1/8. The focus sits inside the parabola on the line of symmetry so that the vertex is midway between the focus and directrix. Since the vertex is at (-1/4,-15/8) and the directrix is f(x)=-1/8, the distance between the vertex and directrix is 15/8-1/8=14/8=7/4 and the focus must be at f(x)=-15/8-7/4=-29/8, making the focus (-1/4,-29/8). All points on the parabola lie equidistant from the directrix and focus. The directrix and focus control the spread of the parabolic arms.  
Read More: ...

ow do you find a and also put in standard form what is the axis of symmetry and what is the y intercept find the zeros and is the parabola upward or downward

If by a you are referring to the form (y-k)=a(x-h)^2. So a=1 and (h,k) is the vertex, with x=h as the line of symmetry. The parabola is like a U, because a>0. (h,k) is (6.36,1.11) and the axis of symmetry is x=6.36. When x=0, y=f(x)=ah^2+k=6.36^2+1.11=41.5596, y intercept. The zeroes are not real, they're complex, because the graph does not cross the x axis (when y=f(x)=0). The minimum value of y is at the vertex. The polynomial can also be written in the form y=ax^2+bx+c; a=1, so y=f(x)=x^2+bx+c=x^2-12.72x+40.4496+1.11=x^2-12.72x+41.5596. The complex zeroes are found by solving (x-6.36)=+sqrt(-1.11)=+isqrt(1.11); x=6.36+isqrt(1.11)=6.36+1.0536i.
Read More: ...

Mechanics projectile M2 question

The trajectory of the particle is a parabola because of gravity. The acceleration due to gravity is 9.81m/s^2 and is, of course, a negative influence on the particle's motion. If we draw a graph of the parabola and make it symmetrical, the point A is on the y axis at y=8, representing 8m. On the x axis at O(-10,0) and P(10,0) the parabola intersects the x axis. The point A(0,8) is the vertex at which the slope is horizontal. Let y=ax^2+bx+c. A is the y intercept so c=8. When y=0, x=10 and -10, so 0=100a+10b+8=100a-10b+8, making b=0, and 100a=-8, so a=-0.08, making the equation of the parabola y=8-0.08x^2. We can relate x and t and y and t, where t is time. The general equation is s=ut-(1/2)gt^2, where s is distance, u an initial speed, and g is the acceleration of gravity which in this case acts against the particle's movement because it is initially projected against gravity. This is a scalar equation, but we need to apply it in a vector situation. Now that we have the equation of the parabola, we need to find out the gradient at O(-10,0) to give us the projection angle. So differentiate y=8-0.08x^2 and we get y'=-0.16x. At x=-10 y'=1.6. The angle of projection z is given by tanz=1.6. Velocity is a vector and the initial projection velocity can be split into a horizontal and vertical component. Gravity acts only on the vertical component. We can apply the motion equation by replacing s with y and u with vsinz, where v is the magnitude of the projection velocity. Therefore y=v(sinz)t-(1/2)gt^2. For x the motion equation is simply x=v(cosz)t.  We can use t to represent the time taken to go from O(-10,0) to P(10,0). The vertical displacement (y) then becomes 0, because the particle starts on the ground and ends on the ground. The horizontal displacement is 20m. By eliminating t between equations we can find the magnitude of v. Putting y=0 in the motion equation: v(sinz)t=(1/2)gt^2, so t=2v(sinz)/g and x=v(cosz)t=2v^2(sinz)(cosz)/g. The horizontal displacement (x) is 20m so 2v^2(sinz)(cosz)/g=20 and v=sqrt(10g/(sinzcosz)). We can calculate sinz and cosz from tanz. sinz=1.6/sqrt(3.56) and cosz=1/sqrt(3.56), so sinzcosz=1.6/3.56. Therefore v=sqrt(98.1*3.56/1.6)=14.774m/s. z, the projection angle=tan^-1(1.6)=58 degrees.  
Read More: ...

what is the axis of symmetry of y=-4(x+8)^2-6

what is the axis of symmetry of y=-4(x+8)^2-6. The right hand side here of this equation is a quadratic and all quadratics are U-shaped. The coefficient of the x^2-term is -4, a negative value, hence the quadratic is a parabola that is U-shaped downwards, like an umbrella. The axis of symmetry is a vertical line that passes through the vertex of the parabola, or U-shape. There are two ways of finding the axis of symmetry. One way is by differentiation to find the maximum (highest) point on the graph (the turning point), which will give the location of the vertex. y = -4(x+8)^2 - 6 dy/dx = -8(x+8) The slope, dy/dx, is zero at x = -8. The turning point, which here is a maximum, occurs at x = -8, so the axis of symmetry is the vertical line, x = -8. Answer: axis of symmetry is x = -8   The other way is by completeing the square, but that is already done in our case, so half our work is already done. We have  y=-4(x+8)^2-6, or (y + 6) = -4(x+8)^2, i.e. Y = AX^2,  where Y = y+6, A = -4, X = x+8. The origin for the X-Y coordinates is X=0, Y=0 and this is the vertex of the parabola and is where the line of symmetry passes through. But X = 0 means x+8 = 0. i.e. x = -8. And Y = 0 means y+6 = 0. i.e. y = -6. So the vertex is (-8, -6) and the axis of symmetry is the vertical line passing through the vertex which is x = -8. Answer: axis of symmetry is x = -8
Read More: ...

How to graph y^2=-12x

This is a sideways parabola, because it is y that is squared. We can write this as x=-(1/12)y^2. Its vertex is at the origin (0,0). The arms of the broad U-shaped parabola point left in the negative region of x. The axis of symmetry is the x axis. The focus is on the axis of symmetry at (0,-3), because if we put a=-(1/12), 1/4a is the distance of the focus from the vertex and 1/4a=-3. The directrix is equidistant outside the parabola at x=3. The latus rectum is where the vertical line through the focus (x=-3) meets the curve: -3=-(1/12)y^2; y=+6 so the end points are (-3,-6) and (-3,6).
Read More: ...

how to graph the parabola for f(x)=x^2-5x-14

f(x)=x^2-5x-14 can be written f(x)=x^2-5x+25/4-25/4-56/4=(x-5/2)^2-81/4. This "completes the square". How did I do this? I looked at x^2-5x and asked myself: what do I need to add to this to complete the square involving x? What you do is halve the x term, 5, and then square it: (5/2)^2=25/4. Now I have the complete square of x-5/2. But I have to compensate for adding in 25/4 so I do this by subtracting -25/4, but keep this separate from the perfect square and absorb it into the number -14. I can represent -14 as an improper fraction -56/4 in a convenient form to combine with -25/4 so that -25/4-56/4=-81/4 is the combined result. Now f(x)=(x-5/2)^2-(9/2)^2, and some of the properties of the parabola can be more easily seen. The graph is U-shaped and cuts the x axis at the zeroes for the function. To find these intercepts, we put (x-5/2)^2-(9/2)^2=0 so (x-5/2)^2=(9/2)^2 and taking square roots of each side we get: x-5/2=+9/2, from which x=5/2+9/2 and x=7 or -2 (also f(x)=(x-7)(x+2)). These are x intercepts and halfway between them at  x=5/2 (1/2 of 7+(-2)) and y=-81/4 is the vertex (in this case the lowest point with x=5/2 as the vertical axis of symmetry). This makes it easy to visualise and draw the graph. We can also work out the y intercept, when x=0. This is -14. All the points labelled fix the symmetrical parabola into position. The arms of the U spread apart and you can plot various values away from the zeroes to see what the spread is.
Read More: ...

i don't understand this exercice

From the first equation, y=x-4, so we substitute this value of y in the second equation: x^2+5y=4 becomes x^2+5(x-4)=4; x^2+5x-20=4; x^2+5x-24=0=(x+8)(x-3)=0. Therefore, x=-8 or 3, making y=x-4=-12 or -1. So the solutions are (x,y)=(-8,-12) or (3,-1). The problem can be solved graphically, too. Draw the graphs of y=x-4 (straight line) and 5y=4-x^2 (parabola) and you'll find the line cuts the parabola at (-8,-12) and (3,-1). The line y=x-4 graph is drawn by marking -4 on the y axis and 4 on the x axis (the intercepts) and drawing a line through these points, extending it beyond these points. The parabola is an inverted U shape. It intersects the x axis at 2 and -2 and its vertex (maximum point) is at y=4/5 the point (0,4/5). You should be able to see how the line intersects the parabola at two points.      
Read More: ...

Tips for a great answer:

- Provide details, support with references or personal experience .
- If you need clarification, ask it in the comment box .
- It's 100% free, no registration required.
next Question || Previos Question
  • Start your question with What, Why, How, When, etc. and end with a "?"
  • Be clear and specific
  • Use proper spelling and grammar
all rights reserved to the respective owners || www.math-problems-solved.com || Terms of Use || Contact || Privacy Policy
Load time: 0.1068 seconds