Let's examine the given function first:
f(x) = x^2 + 1 is the same as f(x) = 1(x-0)^2 + 1.
The vertex of the graph of this function is at (0, 1).
Let x=0 to find the y-intercept: f(0)=0^2+1 = 1; y-int. is at (0,1) (which happens to be the vertex also)
Comparing f(x) = x^2 + 1 to y = x^2, we see that the only difference is that f(x) has a vertical offset of 1. So: Graph y=x^2. Then translate the whole graph UP by 1 unit. That's it. Note (again) that the vertex will be at (0,1), and (0,1) is also the y-intercept.
Answer:
Step-by-step explanation:
ur mom
I need another number for this problem to answer
Step-by-step explanation:
Your assignment probably gave you numbers, like (3,4) or (-5,6). They are graphed on the graph. the first number is where the y axis is and the y axis goes _____ on the coordinate plane. Then he second number is the x axis and that goes up and down like an l. To figure out what the points are you go first in the hose and up the stairs. Put your finger on the dot and count the lines it takes DOWN or UP to get to the middle line. If you ad a negative number you count up to the middle line and positives you count down. Write that number down like this (your 1st number here, Now go back to your point and and count how many lines to get to the middle line that goes up and down otherwise known as your y axis. You either count left or right to get to this line. Count the line it takes to get to that centered line. If you have to count t the right to get this number, mark it as negative. If you count to the left its positive. Now you answer should look like this just replaced with numbers instead of words. (First number here, Second number here) Good luck on your assignment!
