Video: Finding the Intercepts of a Given Function

A function 𝑑(π‘₯) has π‘₯-intercepts at βˆ’4 and 9. Where are the intercepts of 𝑑(π‘₯ βˆ’ 5).

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Video Transcript

A function 𝑑 of π‘₯ has π‘₯-intercepts at negative four and nine. Where are the intercepts of 𝑑 π‘₯ minus five?

𝑑 of π‘₯ has solutions at π‘₯ equals negative four and π‘₯ equals nine. If we substitute π‘₯ minus five in for π‘₯ on both of these equations, we’ll find the new π‘₯-intercepts. π‘₯ minus five equals negative four and π‘₯ minus five equals nine.

To solve for π‘₯, 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. 𝑑 of π‘₯ minus five has intercepts at one and 14. We’ll let the yellow dots represent 𝑑 of π‘₯. And then have the pink representing 𝑑 of π‘₯ minus five.

𝑑 of π‘₯ minus five have intercepts that are five units to the right of the original intercepts. When we subtract from our π‘₯-values, we’re shifting our function to the right to give us our new intercepts at one and 14.

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