Question

a quadratic function $f(x)$ is hidden from view. you must find the $x$-intercept(s) of $f(x)$ and write the answer(s) in the form $(x, y)$. choose the form of the quadratic function $f(x)$ that you would like to see in order to answer the question most efficiently.

form: select a form

answer attempt 1 out of 2

there are no x-intercepts

Explanation:

To find the \( x \)-intercepts of a quadratic function \( f(x) \), the most efficient form of the quadratic function to use is the factored form (also known as the intercept form), which is \( f(x) = a(x - r_1)(x - r_2) \), where \( r_1 \) and \( r_2 \) are the roots (or \( x \)-intercepts) of the quadratic function. Here's why:

Step 1: Understand the Factored Form

The factored form of a quadratic function is \( f(x) = a(x - r_1)(x - r_2) \). In this form, the \( x \)-intercepts (where the graph of the function crosses the \( x \)-axis) occur when \( f(x) = 0 \). Setting \( f(x) = 0 \) gives:

$$ 0 = a(x - r_1)(x - r_2) $$

Since \( a
eq 0 \) (otherwise, it wouldn't be a quadratic function), we can solve for \( x \) by setting each factor equal to zero:

$$ x - r_1 = 0 \quad \text{or} \quad x - r_2 = 0 $$

This gives the solutions \( x = r_1 \) and \( x = r_2 \). The \( x \)-intercepts are then \( (r_1, 0) \) and \( (r_2, 0) \) (since the \( y \)-coordinate at an \( x \)-intercept is always \( 0 \)).

Step 2: Why Factored Form is Efficient

• If we use the standard form \( f(x) = ax^2 + bx + c \), we would need to solve the quadratic equation \( ax^2 + bx + c = 0 \) using methods like factoring, completing the square, or the quadratic formula. This can be more time-consuming, especially if factoring is not straightforward.
• If we use the vertex form \( f(x) = a(x - h)^2 + k \), we would need to convert it to standard form or solve \( a(x - h)^2 + k = 0 \) for \( x \), which also involves additional steps.
• The factored form directly reveals the \( x \)-intercepts (roots) of the quadratic function. Once we have the factored form, we can immediately identify the \( x \)-values where the function crosses the \( x \)-axis by setting each factor equal to zero.

Step 3: Example (if a specific quadratic were given)

Suppose the factored form of a quadratic function is \( f(x) = 2(x - 3)(x + 1) \). To find the \( x \)-intercepts, we set \( f(x) = 0 \):

$$ 0 = 2(x - 3)(x + 1) $$

Since \( 2
eq 0 \), we solve \( x - 3 = 0 \) and \( x + 1 = 0 \):

• \( x - 3 = 0 \) gives \( x = 3 \), so the \( x \)-intercept is \( (3, 0) \).
• \( x + 1 = 0 \) gives \( x = -1 \), so the \( x \)-intercept is \( (-1, 0) \).

Conclusion

To find the \( x \)-intercepts of a quadratic function most efficiently, we should choose the factored form of the quadratic function. This form directly shows the roots ( \( x \)-intercepts) by setting each factor equal to zero, making the process of finding the \( x \)-intercepts straightforward.

So, the most efficient form to choose is the factored form (intercept form) of the quadratic function.

Answer:

To find the \( x \)-intercepts of a quadratic function \( f(x) \), the most efficient form of the quadratic function to use is the factored form (also known as the intercept form), which is \( f(x) = a(x - r_1)(x - r_2) \), where \( r_1 \) and \( r_2 \) are the roots (or \( x \)-intercepts) of the quadratic function. Here's why:

Step 1: Understand the Factored Form

The factored form of a quadratic function is \( f(x) = a(x - r_1)(x - r_2) \). In this form, the \( x \)-intercepts (where the graph of the function crosses the \( x \)-axis) occur when \( f(x) = 0 \). Setting \( f(x) = 0 \) gives:

$$ 0 = a(x - r_1)(x - r_2) $$

Since \( a
eq 0 \) (otherwise, it wouldn't be a quadratic function), we can solve for \( x \) by setting each factor equal to zero:

$$ x - r_1 = 0 \quad \text{or} \quad x - r_2 = 0 $$

This gives the solutions \( x = r_1 \) and \( x = r_2 \). The \( x \)-intercepts are then \( (r_1, 0) \) and \( (r_2, 0) \) (since the \( y \)-coordinate at an \( x \)-intercept is always \( 0 \)).

Step 2: Why Factored Form is Efficient

• If we use the standard form \( f(x) = ax^2 + bx + c \), we would need to solve the quadratic equation \( ax^2 + bx + c = 0 \) using methods like factoring, completing the square, or the quadratic formula. This can be more time-consuming, especially if factoring is not straightforward.
• If we use the vertex form \( f(x) = a(x - h)^2 + k \), we would need to convert it to standard form or solve \( a(x - h)^2 + k = 0 \) for \( x \), which also involves additional steps.
• The factored form directly reveals the \( x \)-intercepts (roots) of the quadratic function. Once we have the factored form, we can immediately identify the \( x \)-values where the function crosses the \( x \)-axis by setting each factor equal to zero.

Step 3: Example (if a specific quadratic were given)

Suppose the factored form of a quadratic function is \( f(x) = 2(x - 3)(x + 1) \). To find the \( x \)-intercepts, we set \( f(x) = 0 \):

$$ 0 = 2(x - 3)(x + 1) $$

Since \( 2
eq 0 \), we solve \( x - 3 = 0 \) and \( x + 1 = 0 \):

• \( x - 3 = 0 \) gives \( x = 3 \), so the \( x \)-intercept is \( (3, 0) \).
• \( x + 1 = 0 \) gives \( x = -1 \), so the \( x \)-intercept is \( (-1, 0) \).

Conclusion

To find the \( x \)-intercepts of a quadratic function most efficiently, we should choose the factored form of the quadratic function. This form directly shows the roots ( \( x \)-intercepts) by setting each factor equal to zero, making the process of finding the \( x \)-intercepts straightforward.

So, the most efficient form to choose is the factored form (intercept form) of the quadratic function.