Problem 34 - College Algebra

Question

Use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range. \(f(x)=6-4 x+x^{2}\)

Step-by-Step Solution

Answer

The vertex of the parabola is at (2,2), the equation for the axis of symmetry is \(x=2\), the parabola has no x-intercept but a y-intercept at (0,6). The domain of the function is all real numbers, and the range is \(y≥2\).

Step 1: Calculate the Vertex

The vertex form of a quadratic is \(y=a(x-h)^2 +k\), where (h,k) are the vertex of the parabola. So first rewrite the standard form \(f(x)=x^2-4 x+6\) to vertex form \(f(x)= (x-2)^2 +2\). Therefore, the vertex of the parabola is (2,2).

Step 2: Calculate the Axis of Symmetry

The axis of symmetry of a parabola is the vertical line through the vertex. It is defined by the equation \(x=h\). Here, as the x-coordinate of the vertex (h.k ) is 2, the equation of the axis of symmetry is given by\(x=2\).

Step 3: Calculate Intercepts

To find the x-intercepts, solve \(f(x)=0\). Which leads to the equation \(x^2-4x+6=0\). Solving this we get \(x=1±i√2\), two complex roots. Hence, it doesn't intersect x-axis. The y-intercept can be found by substituting \(x=0\) in the equation \(f(x)\), this gives \(y=6\). So the y-intercept is (0,6).

Step 4: Graph the Function and Identify Domain and Range

The quadratic function \(f(x)=x^2-4x+6\) is graphed using the vertex at (2,2), the axis of symmetry at \(x=2\) and the y-intercept at (0,6). As for the domain and range: The domain of a function is the set of all possible x-values: As it is a polynomial of degree 2, i.e., a quadratic, the domain, D, in this case, is \(-∞

Key Concepts

Vertex Form

The vertex form of a quadratic function is an elegant way to express quadratic equations, where the form is given by:

• \[ f(x) = a(x-h)^2 + k \]
• Here, \((h, k)\) represents the vertex of the parabola.

When you convert a standard quadratic function like \( f(x) = x^2 - 4x + 6 \) into the vertex form, you essentially open up information about the graph's peak or trough point, known as the vertex.
In this function, rewriting it in vertex form gives us \( f(x) = (x-2)^2 + 2 \), meaning the vertex is located at the point \((2, 2)\).
This place represents the lowest point of the parabola since this particular parabola, indicated by a positive "a" value, opens upwards. With the vertex form, predicting and sketching the graph becomes a lot more intuitive.

Axis of Symmetry

The axis of symmetry is a crucial concept in understanding quadratic functions.
It is a vertical line that bisects the parabola into two mirror images.

• Its equation is \( x = h \), where \( h \) is the x-coordinate of the vertex.

For the function \( f(x) = (x-2)^2 + 2 \), the vertex \((2, 2)\) tells us that the axis of symmetry is the line \( x = 2 \).
This line is significant because any point on the parabola reflected across this line leads to another point on the parabola.
Understanding the axis of symmetry helps in predicting the shape and position of the parabola on the graph.

Intercepts

Intercepts are points where the graph intersects the axes, providing specific clues about the function's behavior.
In our quadratic function, we look for x-intercepts and a y-intercept:

• The x-intercepts are found by solving \( f(x) = 0 \). However, since the roots are complex \( x = 1 \pm i\sqrt{2} \), this parabola doesn't meet the x-axis.
• The y-intercept is identified by setting \( x = 0 \) in the function, which results in \( y = 6 \). Thus, the y-intercept is \((0, 6)\).

These intercepts guide us in painting a more accurate picture of the parabola when sketching it on a coordinate plane.

Domain and Range

When discussing domain and range, it's about exploring the values that x and y can take:

• The domain of any quadratic function, including \( f(x) = x^2 - 4x + 6 \), is all real numbers. This means x can be any value from \(-\infty\) to \(\infty\).
• For the range, since the parabola opens upwards and the vertex \((2, 2)\) is at its minimum, y will be \( 2 \leq y < \infty \).

This information describes all possible y-values that result from plugging x into the function. Knowing the domain and range helps in understanding how the graph behaves and what real-world phenomena it might model.

Previous

Problem 34

Next

Problem 35

Other exercises in this chapter

Problem 34

Find the horizontal asymptote, if any, of the graph of each rational function. $$h(x)=\frac{15 x^{3}}{3 x^{2}+1}$$

View solution

Problem 34

Given \(f(x)=x^{3}-7 x^{2}+5 x-6,\) use the Remainder Theorem to find \(f(3)\).

View solution

Problem 35

In Exercises \(35-50\) a. Use the Leading Coefficient Test to determine the graphs end behavior. b. Find \(x\) -intercepts by setting \(f(x)=0\) and solving the

View solution

Problem 35

In Exercises \(29-36,\) find an nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing utility, use

View solution

Key Concepts

Vertex Form

The vertex form of a quadratic function is an elegant way to express quadratic equations, where the form is given by:

• \[ f(x) = a(x-h)^2 + k \]
• Here, \((h, k)\) represents the vertex of the parabola.

When you convert a standard quadratic function like \( f(x) = x^2 - 4x + 6 \) into the vertex form, you essentially open up information about the graph's peak or trough point, known as the vertex.
In this function, rewriting it in vertex form gives us \( f(x) = (x-2)^2 + 2 \), meaning the vertex is located at the point \((2, 2)\).
This place represents the lowest point of the parabola since this particular parabola, indicated by a positive "a" value, opens upwards. With the vertex form, predicting and sketching the graph becomes a lot more intuitive.

Axis of Symmetry

The axis of symmetry is a crucial concept in understanding quadratic functions.
It is a vertical line that bisects the parabola into two mirror images.

• Its equation is \( x = h \), where \( h \) is the x-coordinate of the vertex.

For the function \( f(x) = (x-2)^2 + 2 \), the vertex \((2, 2)\) tells us that the axis of symmetry is the line \( x = 2 \).
This line is significant because any point on the parabola reflected across this line leads to another point on the parabola.
Understanding the axis of symmetry helps in predicting the shape and position of the parabola on the graph.

Intercepts

Intercepts are points where the graph intersects the axes, providing specific clues about the function's behavior.
In our quadratic function, we look for x-intercepts and a y-intercept:

• The x-intercepts are found by solving \( f(x) = 0 \). However, since the roots are complex \( x = 1 \pm i\sqrt{2} \), this parabola doesn't meet the x-axis.
• The y-intercept is identified by setting \( x = 0 \) in the function, which results in \( y = 6 \). Thus, the y-intercept is \((0, 6)\).

These intercepts guide us in painting a more accurate picture of the parabola when sketching it on a coordinate plane.

Domain and Range

When discussing domain and range, it's about exploring the values that x and y can take:

• The domain of any quadratic function, including \( f(x) = x^2 - 4x + 6 \), is all real numbers. This means x can be any value from \(-\infty\) to \(\infty\).
• For the range, since the parabola opens upwards and the vertex \((2, 2)\) is at its minimum, y will be \( 2 \leq y < \infty \).

This information describes all possible y-values that result from plugging x into the function. Knowing the domain and range helps in understanding how the graph behaves and what real-world phenomena it might model.