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Chapter 10: Problem 27

Sketch the graph of the given equation. $$ 4 x^{2}+16 x-16 y+32=0 $$

Short Answer

Expert verified
The graph is a parabola opening upwards with vertex at (-2, 1) and y-intercept at (0, 2).

Step by step solution

01

Simplify the Equation

First, let's simplify the given quadratic equation so that it can be more recognizable. The given equation is:\[ 4x^2 + 16x - 16y + 32 = 0 \]Let's move \(16y\) to the other side of the equation:\[ 4x^2 + 16x + 32 = 16y \]Now divide all terms by 16 to simplify:\[ \frac{4x^2}{16} + \frac{16x}{16} + \frac{32}{16} = \frac{16y}{16} \]This simplifies to:\[ \frac{x^2}{4} + x + 2 = y \]
02

Rearrange to Standard Parabolic Form

We recognize \(y = \frac{x^2}{4} + x + 2\) as a parabola. Let's rearrange it to reveal the standard form of a parabola \(y = ax^2 + bx + c\):\[ y = \frac{1}{4}x^2 + x + 2 \]
03

Identify the Vertex

The vertex of a parabola in the form \(y = ax^2 + bx + c\) can be found using \(x = -\frac{b}{2a}\). Here, \(a = \frac{1}{4}\) and \(b = 1\). Calculate the x-coordinate of the vertex:\[ x = -\frac{1}{2 \times \frac{1}{4}} = -2 \]Substitute \(x = -2\) back into the equation to find the y-coordinate:\[ y = \frac{1}{4}(-2)^2 + (-2) + 2 = 1 - 2 + 2 = 1 \]Thus, the vertex is \((-2, 1)\).
04

Determine the Direction of the Parabola

Since the coefficient of \(x^2\) is positive \(\left( \frac{1}{4} > 0 \right)\), the parabola opens upwards.
05

Find Y-intercept

Substitute \(x = 0\) into the simplified equation to find the y-intercept:\[ y = \frac{1}{4}(0)^2 + 0 + 2 = 2 \]So, the y-intercept is \((0, 2)\).
06

Sketch the Graph

Using the information, sketch the graph:1. The vertex is at \((-2, 1)\).2. The parabola opens upwards.3. The y-intercept is \((0, 2)\).Plot these points and draw a smooth curve through them to complete the sketch.

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Key Concepts

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

Quadratic Equation
The equation given in the problem, \[ 4x^2 + 16x - 16y + 32 = 0 \] is an example of a quadratic equation. Quadratic equations are polynomials of degree two. They generally take the form \( ax^2 + bx + c = 0\) when expressed as a function in terms of \( x \). This particular form makes it easy to identify the key components, such as the coefficients \( a \), \( b \), and \( c \). The quadratic equation describes a parabola when graphed on a coordinate plane. Key Points about Quadratic Equations:
  • They always contain an \( x^2 \) term, making them non-linear.
  • The coefficient \( a \) determines the opening direction of the parabola (upwards if positive, downwards if negative).
  • Quadratics can be solved using various methods like factoring, completing the square, or using the quadratic formula.
  • Every quadratic equation graphs as a parabolic curve.
Graphing
Graphing a quadratic equation requires identifying its main features. For the equation \( y = \frac{1}{4}x^2 + x + 2 \), you'll see that it represents a parabola.Steps in Graphing a Parabola:
  • Determine the vertex, as it provides the turning point of the parabola.
  • Find the y-intercept, where the parabola crosses the y-axis.
  • Identify the direction in which the parabola opens, determined by the sign of \( a \).
  • Plot several points symmetrically around the vertex for a clear curve outline.
  • Draw the smooth curve, ensuring it passes through the vertex and the intercepts.
Graphing helps visualize and understand the physical properties of quadratic equations.
Vertex of a Parabola
The vertex of a parabola represents its maximum or minimum point. For the equation \( y = ax^2 + bx + c \), the vertex can be calculated using the formula for the x-coordinate: \[ x = -\frac{b}{2a} \] Applying this to \( y = \frac{1}{4}x^2 + x + 2 \), we find the x-coordinate as \( -2 \), giving us a vertex of \((-2, 1)\) after substituting back to find \( y \).Important Features of the Vertex:
  • The vertex can be a point of maximum or minimum height of the parabola, depending on the sign of \( a \).
  • It provides the axis of symmetry for the parabola, where it is symmetric on both sides.
  • Understanding the vertex allows you to identify the initial shape and direction of the parabola.
Y-intercept
The y-intercept of a quadratic equation is the point where the graph crosses the y-axis. This occurs when \( x = 0 \). By substituting \( x = 0 \) into the equation \( y = \frac{1}{4}x^2 + x + 2 \), we find the intercept to be \( y = 2 \), giving us the point \((0, 2)\).Steps to Find the Y-intercept:
  • Set \( x = 0 \) in the quadratic equation.
  • Simplify to solve for \( y \), giving you the y-coordinate of the intercept.
  • Remember, the x-coordinate is always zero for the y-intercept.
The y-intercept is crucial in initial graph plotting, helping you anchor the parabola accurately on the graph.

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Most popular questions from this chapter

Find the equation of the tangent line to the given curve at the given point. \(\frac{x^{2}}{24}+\frac{y^{2}}{16}=1\) at \((3 \sqrt{2},-2)\)

Find the equations of the tangent and the normal lines to the given parabola at the given point. Sketch the parabola, the tangent line, and the normal line. \(x^{2}=-6 y,(3 \sqrt{2},-3)\)

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Name the conic (horizontal ellipse, vertical hyperbola, and so on ) corresponding to the given equation. \(\frac{x^{2}}{9}-\frac{y^{2}}{4}=1\)

The slope of the tangent line to the hyperbola $$ 2 x^{2}-7 y^{2}-35=0 $$ at two points on the hyperbola is \(-\frac{2}{3}\). What are the coordinates of the points of tangency?
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