Problem 11 - Algebra and Trigonometry

Question

\(5-18=\) A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its \(x-\) and \(y\) -intercept(s). (c) Sketch its graph. $$ f(x)=-x^{2}+6 x+4 $$

Step-by-Step Solution

Answer

Vertex at (3, 13); x-intercepts at \(x = 3 \pm \sqrt{13}\); y-intercept at (0, 4); parabola opens downward.

Step 1: Standard Form

The given quadratic function is already in standard form: \( f(x) = -x^2 + 6x + 4 \). The standard form of a quadratic function is \( ax^2 + bx + c \), where \( a = -1 \), \( b = 6 \), and \( c = 4 \).

Step 2: Vertex Calculation

To find the vertex of the quadratic function, use the vertex formula \( x = -\frac{b}{2a} \). Substitute \( b = 6 \) and \( a = -1 \):\[ x = -\frac{6}{2(-1)} = 3.\] Now calculate \( y \) by substituting \( x = 3 \) back into the function:\[ f(3) = -(3)^2 + 6(3) + 4 = -9 + 18 + 4 = 13.\] So, the vertex is \((3, 13)\).

Step 3: X-Intercepts Calculation

To find the x-intercepts, set \( f(x) = 0 \):\[ -x^2 + 6x + 4 = 0. \] Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = -1 \), \( b = 6 \), \( c = 4 \). Calculate the discriminant:\[ b^2 - 4ac = 6^2 - 4(-1)(4) = 36 + 16 = 52.\] Substituting into the quadratic formula, we get:\[ x = \frac{-6 \pm \sqrt{52}}{-2}.\] Simplify \( \sqrt{52} \):\[ x = \frac{-6 \pm 2\sqrt{13}}{-2} = 3 \pm \sqrt{13}.\] So the x-intercepts are \( x = 3 + \sqrt{13} \) and \( x = 3 - \sqrt{13} \).

Step 4: Y-Intercept Calculation

The y-intercept of the function is found by setting \( x = 0 \):\[ f(0) = -(0)^2 + 6(0) + 4 = 4.\] Thus, the y-intercept is \( (0, 4) \).

Step 5: Graph Sketch

To sketch the graph, plot the vertex \((3, 13)\), the y-intercept \((0, 4)\), and the x-intercepts \(x = 3 + \sqrt{13}\) and \(x = 3 - \sqrt{13}\). Since the coefficient of \(x^2\) is negative, the parabola opens downward.

Key Concepts

Vertex

The vertex of a quadratic function is a crucial point on its graph. It represents the maximum or minimum value, depending on the direction in which the parabola opens. For the function given in this exercise, the formula to find the vertex is \( x = -\frac{b}{2a} \), where \( b \) and \( a \) are coefficients from the quadratic equation \( ax^2 + bx + c \).
In our case, \( a = -1 \) and \( b = 6 \). When we substitute these values into the vertex formula, we find \( x = 3 \).
Next, we substitute this \( x \) value back into the original function to find the corresponding \( y \)-coordinate. By calculating \( f(3) \), we get \( y = 13 \).
Thus, the vertex of the quadratic function is at the point \((3, 13)\). It indicates that the parabola reaches its maximum height at this point since the parabola opens downward.

X-intercepts

The \( x \)-intercepts are where the graph of the quadratic function crosses the \( x \)-axis. This is where the function's value is zero. To find these points, you solve the equation \( f(x) = 0 \).
The function given is \( -x^2 + 6x + 4 = 0 \). To solve it, you can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = -1 \), \( b = 6 \), and \( c = 4 \).
Calculate the discriminant \( b^2 - 4ac = 36 + 16 = 52 \). The solutions are:\[ x = \frac{-6 \pm \sqrt{52}}{-2} = 3 \pm \sqrt{13} \]
Therefore, the \( x \)-intercepts are at \( x = 3 + \sqrt{13} \) and \( x = 3 - \sqrt{13} \). These points show where the graph touches or crosses the \( x \)-axis, providing insights into the function's roots.

Y-intercept

The \( y \)-intercept of a function is the point where the graph crosses the \( y \)-axis. It happens when the input \( x \) is zero. To find the \( y \)-intercept of our quadratic function, you simply evaluate \( f(0) \).
In this exercise, substitute \( x = 0 \) into the quadratic function:\[ f(0) = -(0)^2 + 6(0) + 4 = 4 \]Thus, the \( y \)-intercept is \((0, 4)\).
This means that the graph crosses the \( y \)-axis at \( y = 4 \), providing a crucial point to consider when sketching the parabolic shape of the function.

Graph Sketch

Sketching a quadratic function involves plotting key points including the vertex, \( x \)-intercepts, and \( y \)-intercept to draw the general shape of the parabola.

• Start by plotting the vertex \((3, 13)\) on the graph. This point shows the peak because the parabola opens downward.
• Next, include the \( x \)-intercepts, \( x = 3 + \sqrt{13} \) and \( x = 3 - \sqrt{13} \). These are where the graph crosses the \( x \)-axis.
• Also, mark the \( y \)-intercept at \((0, 4)\) on the \( y \)-axis. This helps in shaping the initial direction of the graph.

Since the coefficient of \( x^2 \) in the function is negative, \( -1 \), the parabola opens downwards. Draw a smooth curve through the plotted points that extend downwards in both directions. This sketches a symmetrical shape, tapering off as it moves further from the vertex.