A function 𝑡 of 𝑥 has 𝑥-intercepts at negative four and nine. Where are the intercepts of 𝑡 𝑥 minus five?
𝑡 of 𝑥 has solutions at 𝑥 equals negative four and 𝑥 equals nine. If we substitute 𝑥 minus five in for 𝑥 on both of these equations, we’ll find the new 𝑥-intercepts. 𝑥 minus five equals negative four and 𝑥 minus five equals nine.
To solve for 𝑥, we’ll add five to both sides of the equations, both equations. On the left, negative four plus five equals one. And on the right, nine plus five equals 14. 𝑡 of 𝑥 minus five has intercepts at one and 14. We’ll let the yellow dots represent 𝑡 of 𝑥. And then have the pink representing 𝑡 of 𝑥 minus five.
𝑡 of 𝑥 minus five have intercepts that are five units to the right of the original intercepts. When we subtract from our 𝑥-values, we’re shifting our function to the right to give us our new intercepts at one and 14.