Question

(a) find the vertex and the axis of symmetry of each quadratic function, and determine whether the graph is concave up or concave down. (b) Find the y-intercept and the \(x\) -intercepts, if any. (c) Use parts (a) and (b) to graph the function. (d) Find the domain and the range of the quadratic function. (e) Determine where the quadratic function is increasing and where it is decreasing. (f) Determine where \(f(x)>0\) and where \(f(x)<0\) \(f(x)=3 x^{2}-8 x+2\)

Short answer

Vertex: \( \left( \frac{4}{3}, -\frac{10}{3} \right) \). Axis of Symmetry: \( x = \frac{4}{3} \). Concave up. y-intercept: (0, 2). x-intercepts: \( \frac{4 \pm \sqrt{10}}{3} \). Domain: \( (-\infty, \infty) \). Range: \( [ -\frac{10}{3}, \infty ) \). Decreasing on \( (-\infty, \frac{4}{3}) \). Increasing on \( (\frac{4}{3}, \infty) \).

Step-by-step solution

- Identify Vertex

To find the vertex of the quadratic function in the form of \( f(x) = ax^2 + bx + c \), use the formula for the x-coordinate of the vertex: \( x = -\frac{b}{2a} \). For the function \( f(x) = 3x^2 - 8x + 2 \), we have \(a = 3\), \(b = -8\), and \(c = 2\). Compute \( -\frac{-8}{2(3)} = \frac{8}{6} = \frac{4}{3} \). This is the x-coordinate of the vertex.

- Find y-coordinate of the Vertex

Substitute \( x = \frac{4}{3} \) back into the function to find the y-coordinate: \( f\left( \frac{4}{3} \right) = 3 \left( \frac{4}{3} \right)^2 - 8 \left( \frac{4}{3} \right) + 2 \). Compute this to get \( f\left( \frac{4}{3} \right) = 3 \left( \frac{16}{9} \right) - \frac{32}{3} + 2 \ = \frac{48}{9} - \frac{96}{9} + \frac{18}{9} \ = \frac{-30}{9} = -\frac{10}{3} \). Thus, the vertex is \( \left( \frac{4}{3}, -\frac{10}{3} \right) \).

- Axis of Symmetry

The axis of symmetry for a quadratic function is the vertical line passing through the x-coordinate of the vertex. Therefore, the axis of symmetry is \( x = \frac{4}{3} \).

- Concavity

The concavity of the quadratic function is determined by the coefficient \(a\). If \(a > 0\), the parabola is concave up. If \(a < 0\), the parabola is concave down. Here, \(a = 3 > 0\), so the graph is concave up.

- Find y-intercept

To find the y-intercept, set \(x = 0 \) in the function: \( f(0) = 3(0)^2 - 8(0) + 2 = 2 \). The y-intercept is (0, 2).

- Find x-intercepts

To find the x-intercepts, solve the equation \( 3x^2 - 8x + 2 = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \(a = 3\), \(b = -8\), \(c = 2\). Compute: \( x = \frac{8 \pm \sqrt{64 - 24}}{6} = \frac{8 \pm \sqrt{40}}{6} = \frac{8 \pm 2\sqrt{10}}{6} = \frac{4 \pm \sqrt{10}}{3} \). Thus, the x-intercepts are \( \frac{4 + \sqrt{10}}{3} \) and \( \frac{4 - \sqrt{10}}{3} \).

- Graph the Function

Plot the vertex \( \left( \frac{4}{3}, -\frac{10}{3} \right) \), the y-intercept \( (0, 2) \), and the x-intercepts \( \frac{4 + \sqrt{10}}{3} \) and \( \frac{4 - \sqrt{10}}{3} \). Using the concavity, draw the parabola opening upwards.

- Domain of the Function

The domain of any quadratic function is all real numbers: \( (-\infty, \infty) \).

- Range of the Function

Since the parabola is concave upward, the minimum value is at the vertex. Therefore, the range is \( [ -\frac{10}{3}, \infty ) \).

- Increasing and Decreasing Intervals

The function decreases on the interval \( (-\infty, \frac{4}{3}) \), and increases on the interval \( (\frac{4}{3}, \infty) \).

- Where \( f(x) > 0 \) and \( f(x) < 0 \)

The function \(f(x)\) is greater than 0 when \( x \in (-\infty, \frac{4 - \sqrt{10}}{3}) \cup (\frac{4 + \sqrt{10}}{3}, \infty) \), and less than 0 when \( x \in (\frac{4 - \sqrt{10}}{3}, \frac{4 + \sqrt{10}}{3}) \).

Key concepts

vertex

The vertex of a quadratic function is a crucial point that represents either the highest or the lowest point on the graph, depending on the concavity. To find the vertex, start with the general form of a quadratic function: \(f(x) = ax^2 + bx + c\). Use the formula for the x-coordinate of the vertex: \(x = -\frac{b}{2a}\). For our example \(f(x) = 3x^2 - 8x + 2\), the coefficients are \(a = 3\), \(b = -8\), and \(c = 2\). Plugging these values into the formula gives \(x = \frac{4}{3}\). To get the y-coordinate, substitute \(x = \frac{4}{3}\) back into the function: \(f\big(\frac{4}{3}\big) = 3 \big(\frac{4}{3}\big)^2 - 8 \big(\frac{4}{3}\big) + 2\). Calculate to obtain \(y = -\frac{10}{3}\). Hence, the vertex is \(\big(\frac{4}{3}, -\frac{10}{3}\big)\).

axis of symmetry

The axis of symmetry of a quadratic function is the vertical line that passes through the vertex. It divides the parabola into two mirror-image halves. This line can be found using the x-coordinate of the vertex. For example, using our vertex \(\big(\frac{4}{3}, -\frac{10}{3}\big)\), the axis of symmetry is the line \(x = \frac{4}{3}\). This equation means that for every point to the left of this line, there is an identical point to the right.

concavity

The concavity of a quadratic function tells us whether the parabola opens upward or downward. This is determined by the coefficient \(a\) in the quadratic equation \(f(x) = ax^2 + bx + c\).
If \(a > 0\), the graph is concave up and the parabola opens upwards, resembling a U-shape. If \(a < 0\), the graph is concave down and the parabola opens downwards, resembling an inverted U.
For our function \(f(x) = 3x^2 - 8x + 2\), \(a = 3\), which is greater than 0, so the parabola is concave up.

intercepts

Intercepts are points where the graph crosses the x-axis and y-axis.
**Y-intercept**: To find this, set \(x\) to 0 in the function. For our function:
\(f(0) = 3(0)^2 - 8(0) + 2 = 2\). So, the y-intercept is \((0, 2)\).
**X-intercepts**: These are found by setting \(f(x) = 0\) and solving for \(x\). Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
For our function \(3x^2 - 8x + 2 = 0\), \(a = 3\), \(b = -8\), and \(c = 2\). Calculate:
\(x = \frac{8 \pm \sqrt{64 - 24}}{6} = \frac{8 \pm \sqrt{40}}{6} = \frac{8 \pm 2\sqrt{10}}{6} = \frac{4 \pm \sqrt{10}}{3}\). Thus, the x-intercepts are \(\frac{4 + \sqrt{10}}{3}\) and \(\frac{4 - \sqrt{10}}{3}\).

domain and range

The domain of a quadratic function is all real numbers because you can substitute any real number for \(x\) and get a valid \(y\). For the function \(f(x) = 3x^2 - 8x + 2\), the domain is \( (-\infty, \infty) \).
The range depends on whether the parabola opens upwards or downwards. If it opens upwards, the range is from the y-coordinate of the vertex to infinity. In our case, the vertex is \(\big(\frac{4}{3}, -\frac{10}{3}\big)\), so the range starts from \(-\frac{10}{3}\) and goes to infinity: \( [-\frac{10}{3}, \infty) \).

increasing and decreasing intervals

The intervals where a quadratic function is increasing or decreasing can be determined from the vertex.
A parabola that opens upward (\(a > 0\)) decreases until it reaches the vertex and then starts to increase. For our function with vertex \(\big(\frac{4}{3}, -\frac{10}{3}\big)\), the function decreases on \( (-\infty, \frac{4}{3}) \) and increases on \( (\frac{4}{3}, \infty) \).
If the parabola opened downwards (\(a < 0\)), it would increase until the vertex and then decrease.