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Chapter 2: Problem 39

In Exercises \(35-44,\) graph the quadratic function. $$f(x)=4 x^{2}-5 x+10$$

Short Answer

Expert verified
The parabola with vertex at \( \left( \frac{5}{8}, \frac{135}{16} \right) \) and opens upwards, crosses the \( y \)-axis at \( 10 \).

Step by step solution

01

Identify the Standard Form

The given quadratic function is already in the standard form of a quadratic function, which is \( f(x) = ax^2 + bx + c \). Here, \( a = 4 \), \( b = -5 \), and \( c = 10 \).
02

Find the Vertex

The vertex of a quadratic function \( f(x) = ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). Substitute \( b = -5 \) and \( a = 4 \) into the formula:\[ x = -\frac{-5}{2 \times 4} = \frac{5}{8} \]Now, substitute \( x = \frac{5}{8} \) back into the function to find the \( y \)-coordinate of the vertex:\[ f\left(\frac{5}{8}\right) = 4\left(\frac{5}{8}\right)^2 - 5\left(\frac{5}{8}\right) + 10 \]Simplify to get:\[ f\left(\frac{5}{8}\right) = 4\left(\frac{25}{64}\right) - \frac{25}{8} + 10 = \frac{100}{64} - \frac{200}{64} + \frac{640}{64} = \frac{540}{64} = \frac{135}{16} \]So, the vertex is \( \left( \frac{5}{8}, \frac{135}{16} \right) \).
03

Determine the Axis of Symmetry

The axis of symmetry of a parabola in standard form is the vertical line that passes through the vertex. For our function, this line is \( x = \frac{5}{8} \).
04

Find the Y-Intercept

The \( y \)-intercept of a quadratic function is the point where the graph intersects the \( y \)-axis. This occurs when \( x = 0 \). Substitute \( x = 0 \) into the function:\[ f(0) = 4(0)^2 - 5(0) + 10 = 10 \]Thus, the \( y \)-intercept is \( (0, 10) \).
05

Sketch the Parabola

To sketch the parabola, plot the vertex \( \left( \frac{5}{8}, \frac{135}{16} \right) \) and the \( y \)-intercept \( (0, 10) \). The parabola opens upwards because \( a = 4 > 0 \). Reflect the \( y \)-intercept across the axis of symmetry to find another point: reflect \( (0, 10) \) to \( (\frac{5}{8} \times 2, 10) = (\frac{5}{4}, 10) \). These points and the vertex can guide the shape of the parabola.

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Key Concepts

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

Vertex of a Parabola
The vertex of a parabola is a crucial point as it defines the tip or bottom of the curve, depending on whether the parabola opens upwards or downwards.
In a quadratic function of the form \( f(x) = ax^2 + bx + c \), the formula to find the vertex is given by \( x = -\frac{b}{2a} \). This gives you the x-coordinate of the vertex. To find the y-coordinate, plug this x-value back into the original equation.
Let's apply this to our quadratic function \( f(x) = 4x^2 - 5x + 10 \):
  • Here, \( a = 4 \) and \( b = -5 \).
  • The x-coordinate is \( x = \frac{5}{8} \).
  • Substituting \( x = \frac{5}{8} \) into \( f(x) \) gives us the y-coordinate \( \frac{135}{16} \).
  • The vertex is thus \( \left( \frac{5}{8}, \frac{135}{16} \right) \).
The vertex is particularly useful when sketching the graph of a quadratic function, as it gives you a precise central point which helps guide the shape of the parabola.
Axis of Symmetry
The axis of symmetry of a parabola is an imaginary vertical line that passes through the vertex.
It essentially "divides" the parabola into two mirror-image halves. For any quadratic function of the form \( f(x) = ax^2 + bx + c \), the equation for the axis of symmetry is \( x = -\frac{b}{2a} \).
Let's take another look at our example function:
  • We previously found the x-coordinate of the vertex to be \( \frac{5}{8} \).
  • Therefore, the axis of symmetry is the line \( x = \frac{5}{8} \).
If you are sketching a parabola, you can draw this line to ensure your parabola is symmetric. Points on either side of this line should ideally mirror each other on the graph.
Y-Intercept
The y-intercept of a parabola is the point where the graph crosses the y-axis.
It's important because it gives you a specific point through which the curve passes. You find it by setting the x-value to zero and solving for \( f(x) \).
For our function \( f(x) = 4x^2 - 5x + 10 \):
  • Substitute \( x = 0 \) to find \( f(0) \).
  • This gives you: \( f(0) = 10 \).
  • So the y-intercept is the point \( (0, 10) \).
Plotting this point helps anchor the graph as it provides a definite location on the y-axis that the parabola passes through. It is valuable when drawing the graph and serves as a reference point for other calculations.

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Most popular questions from this chapter

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