Starting with the graph of y=ex , find the equation of tha graph that results fr

Question

Starting with the graph of y=ex , find the equation of tha graph that results from
a) reflecting about the line y=4 b) reflecting about the line x=2 This is part of James Stewart Calculus Early Transcendentals 6E, chapter 1.5 problem 14 but it is explained incorrectly.
Starting with the graph of y=ex , find the equation of tha graph that results from
a) reflecting about the line y=4 b) reflecting about the line x=2 This is part of James Stewart Calculus Early Transcendentals 6E, chapter 1.5 problem 14 but it is explained incorrectly.

Explanation / Answer

first of all, to reflect the graph you have to go from positive to negative. think about looking at yourself in the mirror, its the opposite of what you see. all you need to do to reflect the graph is make it negative. y=-(e^x), f(-x)= e^(-x) a) since the graph of y=e^x crosses the y axis at y=1; P(0,1) you want to move the graph up 3 units. "remember up 3 units" the equation should be y= (e^-x) - 3. remember from equation of a line y=mx + b, b = y value of point in a graph. so by subtracting three, you have to add 3 to make it 0. therefore the graph shifts up 3 units. b) the same principle applies to x, except you keep the change with the x, shifting to the right. y= (e^(-x-2))

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.