Solved on Jan 19, 2024

Solve the quadratic inequality $x^{2} - 4x + 3 \leq 0$ by graphing.

STEP 1

Assumptions
1. We are solving the inequality $x^{2}-4x+3 \leq 0$ .
2. We will graph the corresponding quadratic function $f(x) = x^{2}-4x+3$ .
3. The solutions to the inequality will be the x-values for which the graph of the function is at or below the x-axis.

STEP 2

First, we need to find the roots of the quadratic equation $x^{2}-4x+3 = 0$ since these will be the points where the graph intersects the x-axis.
$x^{2}-4x+3 = (x-1)(x-3) = 0$

STEP 3

Set each factor equal to zero and solve for x to find the roots.
$x-1 = 0 \quad \text{or} \quad x-3 = 0$

STEP 4

Solve the first equation for x.
$x = 1$

STEP 5

Solve the second equation for x.
$x = 3$

STEP 6

Now that we have the roots, we know the graph of $f(x) = x^{2}-4x+3$ will intersect the x-axis at $x=1$ and $x=3$ .

STEP 7

Determine the vertex of the parabola. The x-coordinate of the vertex is given by $-\frac{b}{2a}$ for a quadratic equation in the form $ax^2 + bx + c$ .
$x_{vertex} = -\frac{-4}{2 \cdot 1} = \frac{4}{2} = 2$

STEP 8

Substitute $x_{vertex}$ into the function to find the y-coordinate of the vertex.
$y_{vertex} = (2)^{2} - 4(2) + 3 = 4 - 8 + 3 = -1$

STEP 9

The vertex of the parabola is at the point $(2, -1)$ .

STEP 10

Since the coefficient of $x^2$ is positive, the parabola opens upwards. This means that the graph is below the x-axis between the roots $x=1$ and $x=3$ .

STEP 11

Sketch the graph of the parabola with the roots and vertex labeled. The graph should show that the parabola is below the x-axis for $1 \leq x \leq 3$ .

STEP 12

The solution to the inequality $x^{2}-4x+3 \leq 0$ is the interval where the graph of the function is at or below the x-axis.

STEP 13

Write the solution in interval notation.
$Solution: [1, 3]$
The values of $x$ that satisfy the inequality $x^{2}-4x+3 \leq 0$ are all $x$ in the interval $[1, 3]$ .

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Solve the quadratic inequality $x^{2} - 4x + 3 \leq 0$ by graphing.

STEP 1

Assumptions
1. We are solving the inequality $x^{2}-4x+3 \leq 0$ .
2. We will graph the corresponding quadratic function $f(x) = x^{2}-4x+3$ .
3. The solutions to the inequality will be the x-values for which the graph of the function is at or below the x-axis.

STEP 2

First, we need to find the roots of the quadratic equation $x^{2}-4x+3 = 0$ since these will be the points where the graph intersects the x-axis.
$x^{2}-4x+3 = (x-1)(x-3) = 0$

STEP 3

Set each factor equal to zero and solve for x to find the roots.
$x-1 = 0 \quad \text{or} \quad x-3 = 0$

STEP 4

Solve the first equation for x.
$x = 1$

STEP 5

Solve the second equation for x.
$x = 3$

STEP 6

Now that we have the roots, we know the graph of $f(x) = x^{2}-4x+3$ will intersect the x-axis at $x=1$ and $x=3$ .

STEP 7

Determine the vertex of the parabola. The x-coordinate of the vertex is given by $-\frac{b}{2a}$ for a quadratic equation in the form $ax^2 + bx + c$ .
$x_{vertex} = -\frac{-4}{2 \cdot 1} = \frac{4}{2} = 2$

STEP 8

Substitute $x_{vertex}$ into the function to find the y-coordinate of the vertex.
$y_{vertex} = (2)^{2} - 4(2) + 3 = 4 - 8 + 3 = -1$

STEP 9

The vertex of the parabola is at the point $(2, -1)$ .

STEP 10

Since the coefficient of $x^2$ is positive, the parabola opens upwards. This means that the graph is below the x-axis between the roots $x=1$ and $x=3$ .

STEP 11

Sketch the graph of the parabola with the roots and vertex labeled. The graph should show that the parabola is below the x-axis for $1 \leq x \leq 3$ .

STEP 12

The solution to the inequality $x^{2}-4x+3 \leq 0$ is the interval where the graph of the function is at or below the x-axis.

STEP 13

Write the solution in interval notation.
$Solution: [1, 3]$
The values of $x$ that satisfy the inequality $x^{2}-4x+3 \leq 0$ are all $x$ in the interval $[1, 3]$ .

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Solve the quadratic inequality x2−4x+3≤0x^{2} - 4x + 3 \leq 0x2−4x+3≤0 by graphing.

STEP 1

STEP 2

First, we need to find the roots of the quadratic equation x2−4x+3=0x^{2}-4x+3 = 0x2−4x+3=0 since these will be the points where the graph intersects the x-axis.x2−4x+3=(x−1)(x−3)=0x^{2}-4x+3 = (x-1)(x-3) = 0x2−4x+3=(x−1)(x−3)=0

STEP 3

Set each factor equal to zero and solve for x to find the roots.x−1=0orx−3=0x-1 = 0 \quad \text{or} \quad x-3 = 0x−1=0orx−3=0

STEP 4

Solve the first equation for x.x=1x = 1x=1

STEP 5

Solve the second equation for x.x=3x = 3x=3

STEP 6

Now that we have the roots, we know the graph of f(x)=x2−4x+3f(x) = x^{2}-4x+3f(x)=x2−4x+3 will intersect the x-axis at x=1x=1x=1 and x=3x=3x=3.

STEP 7

Determine the vertex of the parabola. The x-coordinate of the vertex is given by −b2a-\frac{b}{2a}−2ab​ for a quadratic equation in the form ax2+bx+cax^2 + bx + cax2+bx+c.xvertex=−−42⋅1=42=2x_{vertex} = -\frac{-4}{2 \cdot 1} = \frac{4}{2} = 2xvertex​=−2⋅1−4​=24​=2

STEP 8

Substitute xvertexx_{vertex}xvertex​ into the function to find the y-coordinate of the vertex.yvertex=(2)2−4(2)+3=4−8+3=−1y_{vertex} = (2)^{2} - 4(2) + 3 = 4 - 8 + 3 = -1yvertex​=(2)2−4(2)+3=4−8+3=−1

STEP 9

The vertex of the parabola is at the point (2,−1)(2, -1)(2,−1).

STEP 10

Since the coefficient of x2x^2x2 is positive, the parabola opens upwards. This means that the graph is below the x-axis between the roots x=1x=1x=1 and x=3x=3x=3.

STEP 11

Sketch the graph of the parabola with the roots and vertex labeled. The graph should show that the parabola is below the x-axis for 1≤x≤31 \leq x \leq 31≤x≤3.

STEP 12

The solution to the inequality x2−4x+3≤0x^{2}-4x+3 \leq 0x2−4x+3≤0 is the interval where the graph of the function is at or below the x-axis.

STEP 13

Write the solution in interval notation.Solution:[1,3]Solution: [1, 3]Solution:[1,3]The values of xxx that satisfy the inequality x2−4x+3≤0x^{2}-4x+3 \leq 0x2−4x+3≤0 are all xxx in the interval [1,3][1, 3][1,3].

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Solve the quadratic inequality $x^{2} - 4x + 3 \leq 0$ by graphing.

First, we need to find the roots of the quadratic equation $x^{2}-4x+3 = 0$ since these will be the points where the graph intersects the x-axis.
$x^{2}-4x+3 = (x-1)(x-3) = 0$

Set each factor equal to zero and solve for x to find the roots.
$x-1 = 0 \quad \text{or} \quad x-3 = 0$

Solve the first equation for x.
$x = 1$

Solve the second equation for x.
$x = 3$

Now that we have the roots, we know the graph of $f(x) = x^{2}-4x+3$ will intersect the x-axis at $x=1$ and $x=3$ .

Determine the vertex of the parabola. The x-coordinate of the vertex is given by $-\frac{b}{2a}$ for a quadratic equation in the form $ax^2 + bx + c$ .
$x_{vertex} = -\frac{-4}{2 \cdot 1} = \frac{4}{2} = 2$

Substitute $x_{vertex}$ into the function to find the y-coordinate of the vertex.
$y_{vertex} = (2)^{2} - 4(2) + 3 = 4 - 8 + 3 = -1$

The vertex of the parabola is at the point $(2, -1)$ .

Since the coefficient of $x^2$ is positive, the parabola opens upwards. This means that the graph is below the x-axis between the roots $x=1$ and $x=3$ .

Sketch the graph of the parabola with the roots and vertex labeled. The graph should show that the parabola is below the x-axis for $1 \leq x \leq 3$ .

The solution to the inequality $x^{2}-4x+3 \leq 0$ is the interval where the graph of the function is at or below the x-axis.

Write the solution in interval notation.
$Solution: [1, 3]$
The values of $x$ that satisfy the inequality $x^{2}-4x+3 \leq 0$ are all $x$ in the interval $[1, 3]$ .