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Chapter 4: Problem 20

Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts. $$s=-\frac{1}{4} t^{2}+t-1$$

Short Answer

Expert verified
Vertex: (2, 0), Axis: t = 2, Max Value: 0, y-Intercept: (0, -1), x-Intercept: (2, 0)

Step by step solution

01

Identify the coefficients

The quadratic function is given as \( s = -\frac{1}{4} t^{2} + t - 1 \). In this form \( s = at^2 + bt + c \), the coefficients are \( a = -\frac{1}{4} \), \( b = 1 \), and \( c = -1 \).
02

Find the vertex

The vertex \((t, s)\) of a quadratic equation in the form \( at^2 + bt + c \) is found using \( t = -\frac{b}{2a} \). Substitute \( a = -\frac{1}{4} \) and \( b = 1 \) to get \( t = -\frac{1}{2(-\frac{1}{4})} = 2 \). Substitute \( t = 2 \) back into the equation to find \( s \):\[s = -\frac{1}{4}(2)^2 + 2 - 1 = -1 + 2 - 1 = 0.\]Thus, the vertex is \((2, 0)\).
03

Determine the axis of symmetry

The axis of symmetry for a quadratic function \( s = at^2 + bt + c \) is a vertical line given by \( t = -\frac{b}{2a} \) which in this case is \( t = 2 \).
04

Identify the maximum or minimum value

Since the coefficient \( a = -\frac{1}{4} \) is negative, the parabola opens downwards, implying the vertex represents a maximum point. The maximum value of the function is the \( s \)-coordinate of the vertex, which is 0.
05

Find the intercepts

For the y-intercept, set \( t = 0 \):\[s = -\frac{1}{4}(0)^2 + 0 - 1 = -1.\]Thus, the y-intercept is at \((0, -1)\).For the x-intercepts, solve \( -\frac{1}{4} t^2 + t - 1 = 0 \). Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = -\frac{1}{4} \), \( b = 1 \), \( c = -1 \):\[t = \frac{-1 \pm \sqrt{1^2 - 4(-\frac{1}{4})(-1)}}{2(-\frac{1}{4})} = \frac{-1 \pm \sqrt{1 - 1}}{-\frac{1}{2}} = \frac{-1 \pm 0}{-\frac{1}{2}} = 2.\]Thus, the x-intercept is at \((2, 0)\). The vertex \((2, 0)\) is also the x-intercept.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex of a Quadratic Function
In any quadratic function of the form \( s = at^2 + bt + c \), the vertex is a key feature that denotes the peak or the lowest dip of the curve, depending on the function's orientation. To pinpoint the vertex, we turn to the formula \( t = -\frac{b}{2a} \), which gives the \( t \)-coordinate. For our quadratic function \( s = -\frac{1}{4}t^2 + t - 1 \), substituting \( a = -\frac{1}{4} \) and \( b = 1 \) yields \( t = 2 \).
Plugging this value into the function to calculate the \( s \)-coordinate results in \( s = 0 \). Hence, the vertex is at the point \((2, 0)\).
This vital point helps determine the maximum or minimum value the function can reach.
Axis of Symmetry
The axis of symmetry in a quadratic function is an imaginary vertical line that slices the parabola into two mirror-image halves. This line passes right through the vertex, and its equation is given by \( t = -\frac{b}{2a} \). For our specific function, the axis of symmetry is once again \( t = 2 \).
It is vital to emphasize that this line is pivotal for understanding the shape of the parabola, as it also assists in determining other elements like the vertex and the nature of the curve鈥攚hether it opens upwards or downwards.
Maximum and Minimum Values
Quadratic functions may have either a maximum or minimum value based on the coefficient \( a \) of \( t^2 \).
If \( a \) is negative, the parabola arches downwards, signifying a maximum value at the vertex.
If \( a \) is positive, the parabola opens upwards, indicating a minimum value. In our case, \( a = -\frac{1}{4} \), confirming that the parabola is downward-opening.
The maximum value of our function is found at its vertex, \( s = 0 \), at the point \((2, 0)\), which is the highest point on the graph.
X-intercepts and Y-intercepts
Intercepts are points where the graph of the function crosses the \( t \)-axis and \( s \)-axis, providing valuable information about the function. The \( y \)-intercept is found by setting \( t = 0 \), yielding \( s = -1 \). Hence, the \( y \)-intercept is at \((0, -1)\).
To identify the \( x \)-intercepts (roots of the equation), solve \( -\frac{1}{4} t^2 + t - 1 = 0 \). Using the quadratic formula reveals that \( t = 2 \). Therefore, the x-intercept is at \((2, 0)\), which coincides with the vertex.
This dual role underlines the interconnected nature of these points in quadratic functions.

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Most popular questions from this chapter

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