Linear-Quadraric Systems of Equations - Examples, Exercises and Solutions

To solve the problem of finding for which values of $x$, the function $g(x) = -x + 4$ is greater than zero, we begin as follows:

• Step 1: Set up the inequality $g(x) > 0$. This translates to $-x + 4 > 0$.
• Step 2: Solve the inequality:
  • Subtract 4 from both sides: $-x > -4$.
  • Multiply both sides by $-1$, reversing the inequality: $x < 4$.

Therefore, the solution to the problem is that $g(x) > 0$ when $x < 4$.

The corresponding choice that reflects this solution is choice 4: $x < 4$.

To solve the inequality $f(x) = x^2 - 6x + 8 < 0$, we follow these steps:

• Step 1: Find the roots of the equation $f(x) = 0$ using the factoring method.
• Step 2: Determine the intervals formed by these roots.
• Step 3: Test points from each interval to determine where $f(x) < 0$.

Step 1: Factor the quadratic equation $x^2 - 6x + 8 = 0$.
Factoring gives: $(x - 2)(x - 4) = 0$.

Thus, the roots are $x = 2$ and $x = 4$.

Step 2: The roots divide the number line into three intervals: $x < 2$, $2 < x < 4$, and $x > 4$.

Step 3: Choose a test point from each interval and plug it into $f(x)$:

• For $x < 2$, choose $x = 1$: $f(1) = 1^2 - 6(1) + 8 = 1 - 6 + 8 = 3$, which is positive, so $f(x) \geq 0$.
• For $2 < x < 4$, choose $x = 3$: $f(3) = 3^2 - 6(3) + 8 = 9 - 18 + 8 = -1$, which is negative, so $f(x) < 0$.
• For $x > 4$, choose $x = 5$: $f(5) = 5^2 - 6(5) + 8 = 25 - 30 + 8 = 3$, which is positive, so $f(x) \geq 0$.

Therefore, the interval where $f(x) < 0$ is $2 < x < 4$.

The correct choice is:

$2 < x < 4$

To solve this problem, we need to determine whether the provided graph corresponds to a quadratic or linear function.

• First, we observe the shape of the graph.
• The graph shows a downward curve, indicating it is a parabola.
• The general form of a quadratic equation $(y = ax^2 + bx + c)$ implies graph types.
• Let's verify if one of the given quadratic choices represents this graph.

Since the graph is a parabola opening upwards, we'll evaluate the given quadratic equation $y = x^2 - 6x + 8$. Analyzing it and comparing leads to:

• The vertex form can be rewritten or identified mathematically or visually from the given expression.
• This quadratic formula aligns with prominent features of the parabola: its vertex, orientation, and intercepts, matching the graph.

Thus, as the parabola aligns perfectly with quadratic properties such as opening upwards, the formula that describes graph 1 correctly is:
$y = x^2 - 6x + 8$.

To determine for which values of $x$ the condition $f(x) > g(x)$ holds, follow these steps:

• Step 1: Set up the inequality $x^2 - 6x + 8 > -x + 4$.
• Step 2: Rearrange all terms to one side: $x^2 - 6x + 8 + x - 4 > 0$ simplifies to $x^2 - 5x + 4 > 0$.
• Step 3: Solve the related quadratic equation $x^2 - 5x + 4 = 0$. This factors as $(x - 1)(x - 4) = 0$, giving roots $x = 1$ and $x = 4$.
• Step 4: Determine intervals defined by the roots: $(-\infty, 1)$, $(1, 4)$, and $(4, \infty)$.
• Step 5: Test points from each interval in the inequality $x^2 - 5x + 4 > 0$:
  • For $x = 0$ (from interval $(-\infty, 1)$), $x^2 - 5x + 4 = 4$ which is greater than 0.
  • For $x = 2$ (from interval $(1, 4)$), $x^2 - 5x + 4 = -2$ which is less than 0.
  • For $x = 5$ (from interval $(4, \infty)$), $x^2 - 5x + 4 = 9$ which is greater than 0.
• Step 6: From the test points, determine $f(x) > g(x)$ in intervals $(-\infty, 1)$ and $(4, \infty)$.

Thus, $f(x) > g(x)$ for $x < 1$ or $x > 4$.

Therefore, the solution to the problem is $x < 1, 4 < x$, corresponding to choice 2.

To solve the inequality f(x) > 0 , we first need to find the roots of the equation $f(x) = x^2 - 6x + 8$.

1. Find the roots of the quadratic equation:
The quadratic is $x^2 - 6x + 8$. This can be factored into:
$f(x) = (x - 2)(x - 4) = 0$.

2. Calculate the roots:
Setting each factor equal to zero gives the roots $x = 2$ and $x = 4$.

3. Determine the intervals defined by these roots:
The roots divide the x-axis into three intervals: $(-\infty, 2)$, $(2, 4)$, and $(4, \infty)$.

4. Test points in each interval to decide positivity:
- For x < 2 , select $x = 1$: f(1) = 1^2 - 6(1) + 8 = 3 > 0 . Thus, f(x) > 0 in $(-\infty, 2)$.
- For 2 < x < 4 , select $x = 3$: f(3) = 3^2 - 6(3) + 8 = -1 < 0 . Thus, f(x) < 0 in $(2, 4)$.
- For x > 4 , select $x = 5$: f(5) = 5^2 - 6(5) + 8 = 3 > 0 . Thus, f(x) > 0 in $(4, \infty)$.

Therefore, the solution to f(x) > 0 is when x < 2 or x > 4 .

The final solution is: x < 2, 4 < x .