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Chapter 8: Problem 23

For each quadratic function, (a) find the vertex and the axis of symmetry and (b) graph the function. $$h(x)=2 x^{2}-16 x+25$$

Short Answer

Expert verified
The vertex is (4, -7) and the axis of symmetry is x=4.

Step by step solution

01

- Identify coefficients

Recognize the coefficients in the quadratic function of the form \(ax^2 + bx + c\). For the given function \(h(x) = 2x^2 - 16x + 25\), the coefficients are: \(a = 2\), \(b = -16\), and \(c = 25\).
02

- Find the vertex

Use the formula for the x-coordinate of the vertex, which is given by \(x = -\frac{b}{2a}\). Substitute the values of \(a\) and \(b\) into the formula:\[ x = -\frac{-16}{2 \times 2} = \frac{16}{4} = 4 \]Next, substitute \(x = 4\) back into the original function to find the y-coordinate of the vertex:\[ h(4) = 2(4)^2 - 16(4) + 25 = 2(16) - 64 + 25 = 32 - 64 + 25 = -7 \]Thus, the vertex of the function is \((4, -7)\).
03

- Find the axis of symmetry

The axis of symmetry for a quadratic function is a vertical line that passes through the vertex. From the vertex found in the previous step, the axis of symmetry is \(x = 4\).
04

- Graph the function

To graph the function, start by plotting the vertex \((4, -7)\) on the coordinate plane. Next, identify additional points by choosing x-values around the vertex and calculating the corresponding y-values. For example:\(x = 3\), \(h(3) = 2(3)^2 - 16(3) + 25 = 2(9) - 48 + 25 = 18 - 48 + 25 = -5\)\(x = 5\), \(h(5) = 2(5)^2 - 16(5) + 25 = 2(25) - 80 + 25 = 50 - 80 + 25 = -5\)Plot these points \((3, -5)\) and \((5, -5)\), and draw the parabola passing through these points and the vertex, opening upwards because \(a = 2 > 0\).

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Key Concepts

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

Vertex
The vertex of a quadratic function is a crucial point. It reveals the highest or lowest value of the function. We can find the vertex using the formula for the x-coordinate: \(x = -\frac{b}{2a}\). Substituting the values from the given function \(h(x) = 2x^2 - 16x + 25\) gives us: \(x = -\frac{-16}{4} = 4\). To find the y-coordinate, substitute \(x = 4\) back into the equation: \[h(4) = 2(4)^2 - 16(4) + 25 = -7\]. This makes the vertex \((4, -7)\).
Keep in mind:
  • The vertex can be either the highest point if the parabola opens downwards (\(a < 0\)),
  • Or the lowest point if it opens upwards (\(a > 0\)).
Axis of Symmetry
The axis of symmetry is a vertical line that runs through the vertex of a quadratic function. This line divides the parabola into two mirror-image halves.
To find it, we use the x-coordinate of the vertex found earlier. For the function \(h(x) = 2x^2 - 16x + 25\), where the vertex is at (4, -7), the axis of symmetry is \x = 4\.
This knowledge simplifies graphing since values on either side of the axis of symmetry will mirror each other.
Graphing Quadratic Functions
To graph a quadratic function, follow these steps:
  • First, identify and plot the vertex. In our example, the vertex is (4, -7).
  • Next, draw the axis of symmetry, which we've determined to be \x = 4\.
  • Then, choose points to the left and right of the vertex to plot. For \(h(x)\), we calculated \(h(3) = -5\) and \(h(5) = -5\). Plot these points: \( (3, -5) \) and \ (5, -5)\.
  • Finally, draw a smooth curve through these points and the vertex, ensuring the parabola opens upwards, as \(a = 2 > 0\).

This systematic approach helps in visualizing the quadratic function accurately.

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