I need help finding the x intercept of an absolute value equation.

Question

$y= |2x-3| + 2x +6$

Find the $x$ intercept.

(P.S.: In my Algebra teacher's answer document it says that there is no $x$ intercept for this equation. I'm confused as to why that is. I keep getting an $x$ intercept.)

Answer 1

If you substitute $y=0$, you can never have a solution to the equation.
Case 1: Let's say $y=0$. Assume $x>\frac{3}{2}$. The equation reduces to $4x=-3$ which is contradictory because we just assumed that $x$ is positive.
Case 2: Again assume $y=0$. Now take $x<\frac{3}{2}$. The equation reduces to $-2x-3+2x+6=0$ which is even more ridiculous.
Hence, $y$ can never be equal to zero; proving that the function cannot have an $x$-intercept.

Answer 2

This equation can be thought of as the sum of two functions, if you replace y with f(x). The slope, when x is less than 1.5, is equal to 0, and f(0)=9. For x>1.5, the slope is 2, so f(x)=>9 for all x. I think the book meant "Is there an x-intercept to the function?, If so, find it."

Answer 3

$y= |2x-3| + 2x +6$

Find the $x$ intercept.

To solve this problem, will have to remind ourselves that the $x$ intercept will have as coordinates $(x_0,0)$. So, to find $x_0$, will have to solve the equation $y=0$: $$y= \color{blue}{|2x-3|} + \color{black}{2x}+ \color{black}{6}=0\Rightarrow \color{blue}{|2x-3|}=-2x-6 \\\,\\ \therefore\,\,2x-3=\begin{cases}-2x-6 & (1) \\ 2x+6 & (2)\end{cases} \\\,\\ 2x-3=-2x-6\Rightarrow x=3/4$$ We check that value by substitution: $$\require{cancel}2\cdot\dfrac34-3=2\cdot\dfrac34-6 \\ \dfrac32-\dfrac62=\dfrac32-\dfrac{12}2 \\ \color{red}{\cancel{\color{black}{\dfrac32}}}-\dfrac62=\color{red}{\cancel{\color{black}{\dfrac32}}}-\dfrac{12}2 \\ -\dfrac62=-\dfrac{12}2 \\ \text{Contradiction!}$$ So $x=3/4$ is not a solution of $\text{(1)}$.

Let's continue with $\text{(2)}$: $$2x-3=2x+6 \\ \color{red}{\cancel{\color{black}{2x}}} -3=\color{red}{\cancel{\color{black}{2x}}}+6 \\ -3=6\quad\, \\ \text{Contradiction!}$$

So equation $\rm (2)$ admits no solution which means that $y=0$ have no solution.

Therefore, $y=|2x-3|+2x+6$ has no $x$ intercept.

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Exercise:

1. Does $y=|2x-3|+2x+6$ have a $y$ intercept?
2. Does $y=|2x-3|-2x+3$ have an $x$ intercept?
3. Does $y=|2x-3|+2x-3$ have an $x$ intercept?
4. Does $y=\left|\dfrac{3x+2}{x-3}\right|-5$ have an $x$ intercept? Does it have a $y$ intercept?
5. How many $x$ intercepts does $y=|2x-1|-3|2x+4|+7$ have? Does it have any $y$ intercept?