How to find x intercept algebraically?
Understand the Problem
The question is asking how to find the x-intercept of a function algebraically, which typically involves setting the function equal to zero and solving for the variable x.
Answer
The x-intercept of a function is found by solving $f(x) = 0$. For linear functions $x = -\frac{b}{m}$; for quadratic functions $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Answer for screen readers
The x-intercept can be found using the following methods:
For a linear function $f(x) = mx + b$, the x-intercept is:
$$ x = -\frac{b}{m} $$
For a quadratic function $f(x) = ax^2 + bx + c$, the x-intercepts are:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
Steps to Solve
- Set the function equal to zero To find the x-intercept of the function, start by setting the function equal to zero. For example, if the function is $f(x) = ax^2 + bx + c$, set it as:
$$ ax^2 + bx + c = 0 $$
- Rearrange the equation If necessary, rearrange the equation to isolate terms. If it's a quadratic, it is already in the standard form. If it's a linear function like $f(x) = mx + b$, rearranging will look like:
$$ mx + b = 0 $$
- Solve for x Next, solve for $x$. If it's linear, subtract $b$ from both sides and then divide by $m$:
$$ mx = -b \implies x = -\frac{b}{m} $$
If it's quadratic, use the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
- Check your solutions Finally, substitute the x-values back into the original function to ensure they produce a zero output, confirming they are indeed x-intercepts.
The x-intercept can be found using the following methods:
For a linear function $f(x) = mx + b$, the x-intercept is:
$$ x = -\frac{b}{m} $$
For a quadratic function $f(x) = ax^2 + bx + c$, the x-intercepts are:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
More Information
The x-intercept represents the point where the graph of the function crosses the x-axis. It is a vital concept in graphing functions, as it helps in determining where the function equals zero.
Tips
- Forgetting to set the function equal to zero initially.
- Misapplying the quadratic formula or forgetting the plus/minus ($\pm$) when multiple solutions exist.
- Not simplifying the final answer, leaving it in an incomplete form.
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