Guide :

# find the line of symmetry

f(x)=x^2-10x+25 a x=5 b x=-3 c x=3 d x=-5

## Research, Knowledge and Information :

### How to Find Quadratic Line of Symmetry - ThoughtCo

Find Quadratic Line of Symmetry (Kelvinsong/Wikimedia Commons/CC0) A parabola is the graph of a quadratic function. Each parabola has a line of symmetry.

### Finding Lines of Symmetry - Illuminations

Finding Lines of Symmetry. Lesson; 1; 2; 3; 4; 5; 6; ... Also talk about which examples do not have line symmetry and why. This will allow for a rich discussion. Next

### How to Find the Line of Symmetry in a Quadratic Equation ...

How to Find the Line of Symmetry in a Quadratic Equation ... The line of symmetry is an imaginary line which runs down the center of this parabola and cuts it into ...

### How do I find Line of symmetry - Brainly.com

How do I find Line of symmetry 2. Ask for details ; Follow; Report; ... To find the line of symmetry, imagine you are folding the shape, letter, etc. in half.

### Lines of Symmetry of Plane Shapes - Math Is Fun

A Triangle can have 3, or 1 or no lines of symmetry: Equilateral Triangle (all sides equal, all angles equal) ... 3 Lines of Symmetry : 1 Line of Symmetry :

### Find the Axis Of Symmetry: Equation, Formula & Vertex - Study.com

It is the line of symmetry of a parabola and divides a parabola into two equal halves that are ... Find the Axis Of Symmetry: Equation, Formula & Vertex Related Study ...

### Find the line of symmetry for the parabola whose equation is ...

Find the line of symmetry for the parabola whose equation is y = 2x^2 - 4x + 1. - 2495620. 1. Log in ... What is the line of symmetry for the graph of y = -3x^2 ...

### Axis of Symmetry of a Parabola. How to find axis from ...

Every parabola has an axis of symmetry which is the line that runs down its 'center ... There are two different formulas that you can use to find the axis of symmetry.

### Symmetry in Equations - Maths Resources - Math Is Fun

Symmetry in Equations. ... symmetry in an equation are: we understand the equation better; it is easier to plot; it can be easier to solve. When we find a ...

### Identifying the Line of Symmetry: Definition & Examples ...

Identifying the Line of Symmetry Algebraically. ... Identifying the Line of Symmetry: Definition & Examples Related Study Materials. Related; Recently Updated; Popular;

## Suggested Questions And Answer :

### 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

### 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: ...

### Find another such word with at least five letters. vertical line of symmetry.

HAWAII has vertical symmetry.

### 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.

### Parabola of (x+2)^2=-10(y-5) with focus and directrix

When x=-2, y=5. This means the vertex of the parabola is at (-2,5), which is a maximum turning-point meaning that the parabola is an upturned U shape. On either side of the line x=-2 the curve looks the same, as if x=-2 were a mirror, so this is the axis of symmetry and the focus lies on this line. The vertex lies midway between the focus and directrix, and we can call the distance of the vertex from the focus and directrix k, making the line y=5-k the directrix line and (-2,5+k) the focus. All parabolas obey the rule that a point P(x,y) on the curve is equidistant from the directrix line (perpendicular distance) and the focus. The distance of P from the directrix is y-(5-k), and the distance from the focus is sqrt((x-(-2))^2+(y-(5+k))^2) (Pythagoras). These distances must be equal, so (y+k-5)^2=(x+2)^2+(y-k-5)^2. But (x+2)^2=-10(y-5), so (y+k-5)^2=-10(y-5)+(y-k-5)^2, -10(y-5)=(y+k-5+y-k-5)(y+k-5-(y-k-5)) (difference of two squares)=(2y-10)(2k). The factor y-5 drops out: -10=4k, so k=-10/4=-5/2. Therefore the directrix line is y=5+5/2=15/2 and the focus is (-2,5/2).

### 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.

### find the vertex axis of symmetry direction of opening for y=1/32 (x+2)2

The x+2 squared is positive so the opening is as in the arms of a U, directed upwards away from the vertex, which is at (-2,0) because the curve goes no lower than this point. The axis of symmetry is the x value of the vertex because the vertex sits on the axis of symmetry, the vertical line, x=-2.

### how don you graph and find the vertex.....

y=x^2+4x+4=(x+2)^2. a) Vertex is when y is minimum or maximum. The quantity (x+2)^2 is always ≥0 so its minimum value is 0 when x=-2. That makes the vertex (-2,0). b) The vertex is always on the axis of symmetry, which in this case is the line, x=-2. The axis of symmetry is like a mirror reflecting the left and right halves of the parabola. c) Intercepts. When x=0 we have the y intercept y=4; when y=0 we have the x intercept, x=-2. At x=-2 the graph touches the x-axis without cutting or intercepting it. d) Graph:

### Find the point on the curve y=x^2 which is nearest to the point A(0,3)

To find the nearest point on the curve we're looking for the line that is perpendicular to the tangent (normal). First we find the gradient by differentiation: 2x.  The normal has a gradient of -1/(2x) because the product of the slope of the normal and the slope of the tangent (gradient) is -1 and 2x*(-1/(2x))=-1. The gradient of the normal varies according to the position on the curve. The equation of the normal must pass through the point A. So y=mx+b must be satisfied by x=0 and y=3: so b=3 and the equation is y=mx+3, where m is to be determined. This line meets the curve when x^2=mx+3. If we put m=-1/(2x), we get x^2=-1/2+3=5/2 and x=+sqrt(2.5) and y=2.5. Therefore the nearest point is either (sqrt(2.5),2.5) or (-sqrt(2.5),2,5). Since the parabola is symmetrical the points are equidistant from A because A is on the axis of symmetry. The normal at (-sqrt(2.5),2.5) is the reflection of the normal at (sqrt(2.5),2.5). [The equations of the two normals are y=3-x/sqrt(10) (right-hand) and y=3+x/sqrt(10) (left-hand). The distance from A=sqrt(2.75).]