The graph of f(x) = x^2, is reflected in the x axis and shifted to the left 2 units and up 4 units.?
y-4=-(x+2)^2 reflection over x: make it negative - (x+2)^2 shift to the left 2: substitute the value you need into (x-h) where h is the units shifted over. Since h = -2, you replace x with (x- -2) aka (x+2) shift up 4: substitute the value you need into (y-k) where k is the units shifted up. Since k=4, you replace x with (x- 4)
Write the equation of the transformed functions.?
1. y = log (x) y = log (x + 5) ==> translated 5 units left y = log (x + 5) - 1 ==> translated 1 unit down y = -[log (x + 5) - 1] ==> y = 1 - log (x + 5) ==> reflected over x-axis y = 1 - log [7x + 5] ==> horizontally compressed by 7 note that this works with the transforms in the order you gave if the vertical reflection is before the vertical translation, then the "-1" would stay -1 edit: no, a horizontal compression "works backwards": because you multiply by 7 before you take the log, you have to start with something 1/7 as big 1/7 x would be a horizontal stretch by 7 (you divide by 7 before you take the log, so you have to start with something 7 times bigger) ex: if x = 10, then log x = log 10 = 1 if x = 10/7, then log (7x) = log 10 = 1 if x = 70, then log (7x) = ... a much bigger number... this is usually the most confusing transform... *********** 2. m(x) = -2 log (x - 1) + 4 vertical asymptote at x = 1 log functions have no horizontal asymptote 1) shifted 1 to the right vertically stretched by a factor of 2, and reflected over the x-axis shifted up 4
Algebra - write y = x^3 as reflected about the x-axis and the y-axis?
To reflect an equation across the y-axis, substitute -x into x. You get y=(-x)^3 ---> y=-x^3 To reflect an equation across the x-axis, substitute -y into y. You get -y=x^3. That's a perfectly fine equation. However, often it's necessary to isolate y. To do that, just multiply by -1 (move the negative over). You get y=-x^3 By analyzing this a bit further, you can conclude that reflecting y=x^3 across either axis results in the same equation y=-x^3. As for your question, if this is a graded assignment, it's best to go with y=-x^3. If I were the teacher, though, I'd give you credit either way.
How do I reflect a function over a line (graph and equation please)?
By using the equation of the line to substitute f(y) for x and g(x) for y. To learn something you derive the equation yourself and graph it afterwards. Start with something simple like reflecting the ellipsis 4x² + y² = 1 over the line y=x (f(y)= y. g(x)=x) If you can do that try to do it over y=x+1 (f(y)=y-1, g(x)=x+1).After that reflect y=e^x over the line y = 2x-3. If you can do that you have mastered it.Greedz
At which point does the graph of the equation x=-3 intersect the x-axis?
Since it’s x=-3 it is just a vertical line and it’s parallel with the y-axis. It intersects the x-axis at the point (-3,0) and passes through all points of the form (-3,y) where y is any value on the y-axis.
What is 1/X reflected on the Y axis and shifted left?
f(x) = 1/xin order to reflect this function the y- axis you multiply by -1g(x) = -f(x)and to shift the function left: h(x-c) = g(x), h(x) = g(x+c) where c is the number of units the function is shifted left byIin this case I chose c = 3 but it could be any positive number∴ the function 1/x reflected over the y-axis and shifted left 3 units is-1/(x+3)
What would be the equation of the graph [math]y=x^2-1[/math] reflected in the [math]y[/math]-axis?
Let object curve be y=f(x) .The image curve is y=f(-x)=(-x)²-1=x²-1 ,as as the object curve
What is the equation that represents the graph of [math]y = x^3 - x^2 + x - 1[/math] after it is reflected in both the x-axis and y-axis?
What is the equation that represents the graph of [math]y = x^3 - x^2 + x - 1[/math] after it is reflected in both the x-axis and y-axis?The equation? There are many. I don’t know which one you consider to be the equation.Generally, to reflect a graph about the [math]x[/math] axis, simply replace [math]x[/math] with [math]-x[/math]. To reflect a graph about the [math]y[/math] axis, simply replace [math]y[/math] with [math]-y[/math].So to reflect [math]y = x^3-x^2+x-1[/math] about the [math]y[/math] axis, you get [math]-y = x^3 - x^2 + x -1[/math].To reflect that about the [math]x[/math] axis, you get [math]-y = (-x)^3 - (-x)^2 + (-x) -1[/math].That is an equation that meets your requirements. You may wish to simplify it and rearrange it to meet your needs.
Write the transformed function when f(x) = log x is translated 3 units left and stretched vertically by a fact?
f(x) = a * log(x + h) h = +3 a = 2 f(x) = 2 * log(x + 3)
Why does taking the inverse of a function reflect its graph across y=x?
Let us try to answer this question by simultaneously plotting a random function [say: F(x)] and its inverse [let it be: F(x)]. That is, we are drawing the graph of y=F(x) and y=G(x) together on the same graph.Assume a random point, suppose (3,5) is a point on y=F(x). But if (3,5) is a point on y=F(x), that means f(3) = 5. This implies that G(5) = 3 [since G is the inverse]. And so, we would plot (5,3) on y=G(x).Suppose (7, 10) is the next point on y=F(x). Then we would also plot (10, 7) on y=G(x).Notice that the point (10, 7) is the mirror image of the point (7, 10) [when the mirror is kept on the line y=x]. Similarly (3,5) is the mirror image of (5,3) if we use the same mirror. This property is true for any pair of points like these.So whenever you plot a point on y=F(x), you would then plot its mirror image point on y=G(x).Hence, when you’re done plotting all the infinite points, y=G(x) would turn out to be the reflection of y=F(x) about the mirror y=x.