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Chapter 8: Problem 35

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and sketch the graph. See Examples 1 through 4 . $$ f(x)=x^{2}-4 x+5 $$

Short Answer

Expert verified
Vertex is (2, 1). The parabola opens upward. Y-intercept is (0, 5); no real x-intercepts.

Step by step solution

01

Identify the Coefficients

For the quadratic function \( f(x) = ax^2 + bx + c \), identify the coefficients: \( a = 1 \), \( b = -4 \), and \( c = 5 \).
02

Find the Vertex

To find the vertex, use the formula \( x = -\frac{b}{2a} \). Substitute the values: \( x = -\frac{-4}{2(1)} = 2 \). Now calculate \( f(2) = (2)^2 - 4(2) + 5 = 1 \). Therefore, the vertex is at the point \((2, 1)\).
03

Determine the Direction of the Parabola

The parabola opens upward if \( a > 0 \) and downward if \( a < 0 \). Here, \( a = 1 \), which is positive, so the parabola opens upward.
04

Find the Y-intercept

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = 0^2 - 4(0) + 5 = 5 \). Thus, the y-intercept is \( (0, 5) \).
05

Determine the X-intercepts

Find the x-intercepts by solving \( x^2 - 4x + 5 = 0 \). Calculate the discriminant: \( b^2 - 4ac = (-4)^2 - 4(1)(5) = 16 - 20 = -4 \), which is negative. Thus, there are no real x-intercepts.
06

Sketch the Graph

Plot the vertex \((2, 1)\) and the y-intercept \((0, 5)\) on the graph. Since the parabola opens upward and has no x-intercepts, sketch the parabola opening up with its vertex as the lowest point.

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Key Concepts

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

Vertex of a Parabola
A quadratic function is expressed in the form \( f(x) = ax^2 + bx + c \). The vertex of a parabola, a key feature in its graph, is essentially its tip. This can be either the highest or lowest point depending on the orientation of the parabola. In our function \( f(x) = x^2 - 4x + 5 \), we identify the vertex using the formula for the x-coordinate: \( x = -\frac{b}{2a} \), where \( a = 1 \), \( b = -4 \). This gives us \( x = -\frac{-4}{2 \times 1} = 2 \). With the x-coordinate in hand, we can substitute it back into the original equation to find the y-coordinate: \( f(2) = 2^2 - 4 \times 2 + 5 = 1 \). Hence, the vertex is \((2, 1)\). Always remember:
  • The vertex represents a critical turning point on the graph.
  • For upward-opening parabolas, it is the lowest point.
  • For downward-opening ones, it is the highest point.
Discriminant
The discriminant is a special magic number that tells us about the nature of the roots of a quadratic equation. It is found using the formula \( b^2 - 4ac \). This handy little tool determines how many and what type of solutions you'll find from your quadratic equation. In our example with \( f(x) = x^2 - 4x + 5 \), we calculate the discriminant as follows:
  • \( b = -4 \)
  • \( a = 1 \)
  • \( c = 5 \)
Thus, \( b^2 - 4ac = (-4)^2 - 4 \times 1 \times 5 = 16 - 20 = -4 \). Since the discriminant is negative, we know:
  • The quadratic equation has no real roots.
  • There are no x-intercepts on the graph.
The discriminant is crucial in indicating the existence of real roots and their nature without needing to graph the function.
X-intercepts
The x-intercepts (also known as the roots or zeros of the function) are the points where the graph of the function crosses the x-axis. This happens when \( f(x) = 0 \). In our case, solving \( x^2 - 4x + 5 = 0 \) reveals whether or not x-intercepts exist.
  • The discriminant, \( b^2 - 4ac \), is negative (-4).
  • When the discriminant is negative, it means the quadratic equation has no real solutions.
Thus, there are no x-intercepts for this function. This indicates our parabola, which opens upwards, stays above the x-axis and never touches it.
Y-intercept
The y-intercept is the point where the graph of the quadratic function intersects the y-axis. This occurs when \( x = 0 \). To find the y-intercept for our function \( f(x) = x^2 - 4x + 5 \), we simply substitute zero into the function for \( x \):
  • \( f(0) = 0^2 - 4 \times 0 + 5 = 5 \)
Therefore, the y-intercept is at the point \((0, 5)\). This point is where the graph will intersect the y-axis, and knowing it helps you begin sketching the graph or predicting its behavior. The intercept acts as an anchor point often used with the vertex to outline the rough shape of the parabola before plotting other accurate points if necessary.

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Most popular questions from this chapter

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