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Chapter 41: Problem 877

Construct the graph of the function defined by \(y=x^{2}-6 x+10\).

Short Answer

Expert verified
The graph of the function \(y = x^2 - 6x + 10\) can be constructed by first finding the vertex at the point (3, 1), the axis of symmetry at \(x = 3\), and the y-intercept at the point (0, 10). Plot these key points on a coordinate plane and sketch the parabola, which opens upward and is symmetric about the axis of symmetry. There are no real x-intercepts in this case.

Step by step solution

01

Identify the Vertex

To find the vertex of the parabola, we use the formula \(x_v = \frac{-b}{2a}\), where \(a\) is the coefficient of the \(x^2\) term and \(b\) is the coefficient of the \(x\) term. In this case, \(a = 1\) and \(b = -6\). So, we have: \(x_v = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3\) Now, substitute \(x_v\) back into the original function to find the corresponding y-coordinate of the vertex: \(y_v = (3)^2 - 6(3) + 10 = 9 - 18 + 10 = 1\) So, the vertex of the parabola is at the point (3, 1).
02

Identify the Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex and divides the parabola into two equal halves. For a quadratic function, the equation of the axis of symmetry is always of the form \(x = x_v\). In this case, \(x_v = 3\), so the axis of symmetry is \(x = 3\).
03

Identify the Intercepts

1. To find the x-intercepts (if any), set \(y = 0\) and solve for \(x\): \(0 = x^2 - 6x + 10\) This equation does not factor easily, so we will use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = 10\). Applying the formula: \(x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(10)}}{2(1)} = \frac{6 \pm \sqrt{36 - 40}}{2}\) Since the discriminant is negative (\(36 - 40 = -4\)), there are no real x-intercepts for this function. 2. To find the y-intercept, set \(x = 0\) and solve for \(y\): \(y = (0)^2 - 6(0) + 10 = 10\) Thus, the y-intercept is at the point (0, 10).
04

Plot the Key Points and Sketch the Graph

Now we will plot the vertex (3, 1), the axis of symmetry (x = 3), and the y-intercept (0, 10) on a coordinate plane. Based on these key points, we can sketch the parabola by drawing a smooth curve that opens upward (since \(a > 0\)). The parabola should be symmetric about the axis of symmetry. With those points plotted and the curve drawn, we have successfully constructed the graph of the function \(y = x^2 - 6x + 10\).

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Key Concepts

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

Vertex of a Parabola
In a quadratic function, the vertex is the peak or the lowest point of the parabola, serving as a critical characteristic. For a function in standard form, such as \( y = ax^2 + bx + c \), the vertex
  • represents the maximum or minimum value of the quadratic function.
  • can be identified using the formula: \( x_v = \frac{-b}{2a} \).
  • when found, the x-value gives the line of symmetry, and substituting it back into the equation provides the y-coordinate of the vertex.
In our given problem, for \(y = x^2 - 6x + 10\), we used \( a = 1 \) and \( b = -6 \). After applying the formula, the x-coordinate of the vertex was found to be 3. Plugging \( x_v = 3 \) back into the function calculates the y-coordinate to form the vertex at (3, 1). Understanding the vertex helps in sketching the overall curve of the parabola.
Axis of Symmetry
The axis of symmetry of a parabola is a vertical line that divides the parabola into two symmetrical halves. It passes through the vertex of the parabola. The equation of this line can quickly be determined once you have the x-coordinate of the vertex from the formula \( x = \frac{-b}{2a} \):
  • This line ensures that each point on one side of the parabola has a mirror image on the opposite side.
  • For the function \( y = x^2 - 6x + 10 \), with vertex x-coordinate found at 3, the axis of symmetry is \( x = 3 \).
Recognizing this axis helps in accurately graphing the parabola. It provides a guideline to ensure that the curve is balanced evenly on both sides of this vertical line.
Quadratic Formula
The quadratic formula is a powerful tool for finding the roots of any quadratic equation like \( ax^2 + bx + c = 0 \), even when they do not factor neatly:
  • The formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • The discriminant \( b^2 - 4ac \) guides us: if it's positive, there are two real roots; if zero, one real root; if negative, no real roots.
  • For \( y = x^2 - 6x + 10 \), the discriminant was negative (\(-4\)), indicating no real x-intercepts.
Understanding negative discriminants is crucial. For this function, the graph does not cross the x-axis, affirming no real roots. This quirk is important when considering the full graph of the quadratic function.
Y-Intercept of a Function
The y-intercept in any function is the point where the graph crosses the y-axis, giving insight into the behavior of the equation at this intersection. For quadratics like \( y = ax^2 + bx + c \):
  • Set \( x = 0 \) to find \( y \), simplifying the graph's representation.
  • For \( y = x^2 - 6x + 10 \), the calculation gives \( y = 10 \).
  • The y-intercept here is at the point (0, 10), serving as a key plot point while sketching the parabola.
Clearly marked, the y-intercept guides in setting the initial position of the parabola on a graph. While it doesn't affect the shape, it marks critical positioning details on the coordinate system.

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Most popular questions from this chapter

Discuss the graph of the parabola \((\mathrm{x}-\mathrm{a})^{2}=4 \mathrm{p}(\mathrm{y}-\mathrm{b})\), and find its axis, focus, directrix, vertex and latus rectum.

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Discuss the graph of the equation \(y^{2}=12 \mathrm{x}\).

Find, both analytically and graphically, the points of intersection of the two curves whose equations are $$ 2 x+y-4=0 \text { and } y^{2}-4 x=0 $$
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