Find the X-Intercepts: Solving the Quadratic Equation x² - 6x + 5

Q: How do I know which two numbers to use for factoring?

Look for two numbers that multiply to give you the constant term (5) and add to give you the middle coefficient (-6). Try different factor pairs: 1×5=5, but 1+5=6 (wrong sign), so use -1 and -5.

Q: What if the quadratic doesn't factor nicely?

Use the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $. This works for any quadratic equation, even when factoring is difficult or impossible.

Q: Can x-intercepts be negative numbers?

Absolutely! X-intercepts can be any real numbers - positive, negative, or zero. In this problem, both intercepts (1 and 5) happen to be positive, but that's not always the case.

Q: How do I write the final answer correctly?

X-intercepts are points on the coordinate plane, so write them as ordered pairs: (x, 0). Since y = 0 at x-intercepts, points A and B are (1, 0) and (5, 0).

Quadratic Factoring with X-Intercepts

The following function has been graphed below:

$f(x)=x^2-6x+5$

Calculate points A and B.

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the coordinates of points A,B

00:03 Notice that points A,B are the intersection points with the X-axis

00:07 At the intersection points with the X-axis, the Y value must = 0

00:15 Substitute Y = 0 and solve for X values

00:21 Break down the function into a trinomial

00:28 This is the corresponding trinomial

00:32 Find what zeros each factor in the product

00:36 This is one solution

00:40 This is the second solution

00:46 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

The following function has been graphed below:

$f(x)=x^2-6x+5$

Calculate points A and B.

Step-by-step solution

To solve for the points A and B, we need to find the roots of the function $f(x) = x^2 - 6x + 5$ where $f(x) = 0$ .

Let's proceed step-by-step:

Step 1: Set the function to zero
We begin by setting the equation to zero: $x^2 - 6x + 5 = 0$ .
Step 2: Factor the quadratic
We need to factor the expression. We look for two numbers that multiply to $c = 5$ and add to $b = -6$ . These numbers are $-1$ and $-5$ .
Step 3: Write the factorization
Therefore, we can write the quadratic as: $(x - 1)(x - 5) = 0$ .
Step 4: Solve for the roots
Set each factor equal to zero: \begin{align*} x - 1 &= 0 \\ x &= 1 \end{align*} \begin{align*} x - 5 &= 0 \\ x &= 5 \end{align*} The roots are $x = 1$ and $x = 5$ .
Step 5: Identify the Points A and B
The points A and B, where the function intersects the x-axis, are $(1, 0)$ and $(5, 0)$ .

Thus, the coordinates of points A and B are $(1,0),(5,0)$ , which matches choice 1.

Final Answer

$(1,0),(5,0)$

Key Points to Remember

Essential concepts to master this topic

X-Intercepts: Set f(x) = 0 to find where parabola crosses x-axis
Factoring: Find two numbers that multiply to 5 and add to -6: (-1)(-5) = 5, -1 + (-5) = -6
Verification: Check solutions: $1^2 - 6(1) + 5 = 0$ and $5^2 - 6(5) + 5 = 0$ ✓

Common Mistakes

Avoid these frequent errors

Confusing x-intercepts with y-intercepts
Don't find where x = 0 when looking for points A and B = wrong axis intersection! This gives you the y-intercept (0, 5) instead of x-intercepts. Always set f(x) = 0 to find x-intercepts where the graph crosses the x-axis.

Practice Quiz

Test your knowledge with interactive questions

The following function has been graphed below:

$$ f(x)=-x^2+5x+6 $$

Calculate points A and B.

FAQ

Everything you need to know about this question

Why do I set the function equal to zero?

X-intercepts occur where the graph crosses the x-axis, meaning the y-coordinate is zero. Since f(x) represents the y-value, setting $f(x) = 0$ finds these crossing points!

How do I know which two numbers to use for factoring?

Look for two numbers that multiply to give you the constant term (5) and add to give you the middle coefficient (-6). Try different factor pairs: 1×5=5, but 1+5=6 (wrong sign), so use -1 and -5.

What if the quadratic doesn't factor nicely?

Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . This works for any quadratic equation, even when factoring is difficult or impossible.

Can x-intercepts be negative numbers?

Absolutely! X-intercepts can be any real numbers - positive, negative, or zero. In this problem, both intercepts (1 and 5) happen to be positive, but that's not always the case.

How do I write the final answer correctly?

X-intercepts are points on the coordinate plane, so write them as ordered pairs: (x, 0). Since y = 0 at x-intercepts, points A and B are (1, 0) and (5, 0).

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