How to reflect a graph over the y axis?
We can reflect the graph of any function f about the x-axis by graphing y=-f(x) and we can reflect it about the y-axis by graphing y=f(-x). We can even reflect it about both axes by graphing y=-f(-x). See how this is applied to solve various problems.
In this regard, how do you reflect a graph on the y axis?Reflection across the y-axis: y = f ( − x ) y = f(-x) y=f(−x) Besides translations, another kind of transformation of function is called reflection. If a reflection is about the y-axis, then, the points on the right side of the y-axis gets to the right side of the y-axis, and vice versa.
Bearing in mind, how do you show a reflection over the y axis in an equation?To flip or reflect (horizontally) about the vertical y-axis, replace y = f(x) with y = f(-x).
In addition how do you find the Y intercept of a reflection?To find the reflection of the y intercept, duplicate the y value of the point and find the x distance to the AOS then travel the same distance on the other side of the AOS. In this case, the y value of the reflection of the y intercept, (0, -1) is –1, so the reflected point will also have a y value of –1.
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Related questions and answers
A parabola can have 2 x-intercepts, 1 x-intercept or zero real x intercepts. If the parabola only has 1 x-intercept (see middle of picture below), then the parabola is said to be tangent to the x-axis.
The location of a stationary point on f(x) can be identified by solving f'(x) = 0. To work out which is the minimum and maximum, differentiate again to find f''(x). Input the x value for each turning point. If f''(x) > 0 the point is a minimum, and if f''(x) < 0, it is a maximum.
Use the symmetry of the graph to find the coordinates of the turning point of the following quadratic:
- y = x 2 − 3 x + 2 y=x^2-3x+2 y=x2−3x+2.
- x 2 − 3 x + 2 = ( x − 1 ) ( x − 2 ) x^2-3x+2=(x-1)(x-2) x2−3x+2=(x−1)(x−2)
- ( x − 1 ) ( x − 2 ) = 0 (x-1)(x-2)=0 (x−1)(x−2)=0.
A local maximum point on a function is a point (x,y) on the graph of the function whose y coordinate is larger than all other y coordinates on the graph at points "close to'' (x,y). Similarly, (x,y) is a local minimum point if it has locally the smallest y coordinate.
A maximum turning point is a turning point where the curve is concave up (from increasing to decreasing ) and f′(x)=0 f ′ ( x ) = 0 at the point.
Relative mins are the lowest points in their little neighborhoods. f has a relative min of -3 at x = -1. f has a relative min of -1 at x = 4.
First, identify the leading term of the polynomial function if the function were expanded. Then, identify the degree of the polynomial function. This polynomial function is of degree 4. The maximum number of turning points is 4 – 1 = 3.
The minimum value of a function is the place where the graph has a vertex at its lowest point.
The y- coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum, respectively. To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function. See Figure 10.
We've learned that a quadratic equation is an equation of degree 2. The standard form of a quadratic is y = ax^2 + bx + c, where a, b, and c are numbers and a cannot be 0. All quadratic equations graph into a curve of some kind. All quadratics will have two solutions, but not all may be real solutions.
One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value.
Just like looking at a mirror image of yourself, but flipped.a reflection point is the mirror point on the opposite side of the axis.
When a function's slope is zero at x, and the second derivative at x is:
- less than 0, it is a local maximum.
- greater than 0, it is a local minimum.
- equal to 0, then the test fails (there may be other ways of finding out though)
This means that relative extrema do not occur at the end points of a domain. They can only occur interior to the domain. Both of these points are relative maximums since they are interior to the domain shown and are the largest point on the graph in some interval around the point.
A reflection is the flipping of a point or figure over a line of reflection (called the mirror line), and a quadratic equation is an equation where the highest exponent is 2. When reflecting over the x-axis, the function f(x) becomes -f(x). For instance, y = 3x^2 would become y = -(3x^2).
If you have the graph, or can draw the graph, the maximum is just the y value at the vertex of the graph. If you are unable to draw a graph, there are formulas you can use to find the maximum. If you are given the formula y = ax2 + bx + c, then you can find the maximum value using the formula max = c - (b2 / 4a).
A relative maximum or minimum occurs at turning points on the curve where as the absolute minimum and maximum are the appropriate values over the entire domain of the function. In other words the absolute minimum and maximum are bounded by the domain of the function.
The easiest way to find the turning point is when the quadratic is in turning point form (y = a(x - h)2 + k), where (h, k) is the turning point. To get a quadratic into turning point form you need to complete the square.
Finding relative maxima and minima of a function can be done by looking at a graph of the function. A relative maximum is a point that is higher than the points directly beside it on both sides, and a relative minimum is a point that is lower than the points directly beside it on both sides.
The graph of a quadratic function is a parabola whose axis of symmetry is parallel to the y -axis. The coefficients a,b, and c in the equation y=ax2+bx+c y = a x 2 + b x + c control various facets of what the parabola looks like when graphed.
Find the first derivative of a function f(x) and find the critical numbers. Then, find the second derivative of a function f(x) and put the critical numbers. If the value is negative, the function has relative maxima at that point, if the value is positive, the function has relative maxima at that point.