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 \(f(x)<0\) \(f(x)=-3 x^{2}+3 x-2\)

Short answer

Vertex: (\( \frac{1}{2}, -0.625 \)), Axis of symmetry: \( x=\frac{1}{2} \), Concave down, y-intercept: (0, -2), No x-intercepts, Domain: \( -\infty <x< \infty \), Range: \( -\infty < y \leq -0.625 \), Increasing: \( -\infty <x< \frac{1}{2} \), Decreasing: \( \frac{1}{2} <x< \infty \)

Step-by-step solution

Find the vertex

The vertex of a quadratic function in the form of \( f(x) = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex, \( x = \frac{-b}{2a} \). For \( f(x) = -3x^2 + 3x - 2 \), a = -3, b = 3. x = \frac{-3}{2(-3)} \ = \frac{-3}{-6} \ = \frac{1}{2}. To find the y-coordinate, substitute \( x = \frac{1}{2} \) back into the function: \( f(x) = -3(\frac{1}{2})^2 + 3(\frac{1}{2}) - 2 \ = \frac{3}{4} + \frac{3}{2} - 2 \ = 0.625 \) So, the vertex is \( (\frac{1}{2}, -0.625) \).

Find the axis of symmetry

The axis of symmetry can be found using the vertex x-coordinate: \( x = \frac{-b}{2a} \ = \frac{1}{2} \).

Determine if graph is concave up or concave down

Since the coefficient of \( x^2 \) is \( -3 \), which is less than 0, the parabola opens downwards (concave down).

Find the y-intercept

The y-intercept is the value of the function when \( x = 0 \): \( f(0) = -3(0^2) + 3(0) - 2 = -2 \). So the y-intercept is \( (0, -2) \).

Find the x-intercepts

The x-intercepts are found by setting \( f(x) = 0 \) and solving for \( x \): \( -3x^2 + 3x - 2 = 0 \). Using the quadratic formula \( x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \), \( a = -3 \), \( b = 3 \), \( c = -2 \) results in: \[ x = \frac{-3 \pm \sqrt{3^2 - 4(-3)(-2)}}{2(-3)} \ = \frac{-3 \pm \sqrt{9 - 24}}{-6} \ = \frac{-3 \pm \sqrt{-15}}{-6} \]. Since there is a negative number inside the square root, there are no real x-intercepts.

Graph the function

Using the vertex \( (\frac{1}{2}, -0.625) \), the axis of symmetry \( x = \frac{1}{2} \), and the y-intercept \( (0, -2) \), draw the parabola opening downwards.

Find the domain

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

Find the range

The range can be found since the parabola opens downwards and the maximum value is at the vertex: \( -\infty < y \leq -0.625 \).

Determine where the function is increasing or decreasing

The function increases on the interval \( -\infty < x < \frac{1}{2} \) and decreases on the interval \( \frac{1}{2} < x < \infty \).

Determine where \( f(x) > 0 \) and \( f(x) < 0 \)

Since the quadratic function is concave down and the vertex is the maximum point, \( f(x) > 0 \) in the intervals \( x \in (-\infty, \frac{1}{2}) \) and \( f(x) < 0 \) in the intervals \( x \in (\frac{1}{2}, \infty) \).

Key concepts

vertex

The vertex of a quadratic function is a point where the curve changes direction. This point can either be a maximum or a minimum. For a function in the form of \(f(x) = ax^2 + bx + c\), the vertex can be found using the formula for the x-coordinate: \( x = \frac{-b}{2a} \). Once you have the x-coordinate, substitute it back into the original equation to find the y-coordinate. For example, in the function \( f(x) = -3x^2 + 3x - 2 \), the vertex is located at \((\frac{1}{2}, -0.625)\).

axis of symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola. It divides the parabola into two mirror images. For the quadratic function in standard form \(f(x) = ax^2 + bx + c\), the axis of symmetry can also be found using the formula \( x = \frac{-b}{2a} \). For instance, in the function \( f(x) = -3x^2 + 3x - 2 \), the axis of symmetry is \( x = \frac{1}{2} \).

concavity

Concavity describes the direction in which a parabola opens. If the coefficient of the \(x^2\) term, denoted as \(a\), is positive, the parabola opens upwards (concave up). If \(a\) is negative, the parabola opens downwards (concave down). For the function \( f(x) = -3x^2 + 3x - 2 \), \( a = -3 \), which is less than 0, hence the graph is concave down.

y-intercept

The y-intercept is the point where the graph of the function crosses the y-axis. It can be found by evaluating the function at \(x = 0\). For the quadratic function \( f(x) = -3x^2 + 3x - 2 \), substituting \(x = 0\) results in \( f(0) = -2 \). Therefore, the y-intercept is at the point \((0, -2)\).

x-intercepts

The x-intercepts are the points where the graph of the function crosses the x-axis. These points can be found by setting the function \(f(x)\) equal to zero and solving for \(x\). In our example, solving \( -3x^2 + 3x - 2 = 0 \) using the quadratic formula \( x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \) results in no real solutions because the discriminant is negative (inside the square root yields a negative number). Thus, there are no real x-intercepts.

domain and range

The domain of a quadratic function is all possible x-values, which is always all real numbers \( -\infty < x < \infty \). The range of a quadratic function depends on whether it opens upwards or downwards. Since the function \( f(x) = -3x^2 + 3x - 2 \) opens downwards with a vertex at \((\frac{1}{2}, -0.625)\), the range is all y-values less than or equal to \(-0.625\): \( -\infty < y \leq -0.625 \).

increasing and decreasing intervals

A function increases on an interval where its graph rises as \(x\) increases, and decreases on an interval where its graph falls. For the quadratic function \( f(x) = -3x^2 + 3x - 2 \), the interval where the function is increasing is \( -\infty < x < \frac{1}{2} \) since it rises toward the vertex. After reaching the vertex, it decreases on the interval \( \frac{1}{2} < x < \infty \).

function positivity and negativity

To determine where a quadratic function is positive (where \( f(x) > 0 \)) and negative (where \( f(x) < 0 \)), observe its vertex and concavity. For \( f(x) = -3x^2 + 3x - 2 \), the maximum point is at the vertex. Since the function is concave down and the vertex is at \( x = \frac{1}{2} \), it will be positive where \( x < \frac{1}{2} \) and negative after \( x > \frac{1}{2} \).