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Chapter 2: Problem 30

Find the intercepts and graph each equation by plotting points. Be sure to label the intercepts. $$ 4 x^{2}+y=4 $$

Short Answer

Expert verified
Y-intercept: \( (0, 4) \); X-intercepts: \( (1, 0) \) and \(( -1, 0)\).

Step by step solution

01

Find the y-intercept

To find the y-intercept, set \(x = 0\). The equation becomes \(4(0)^2 + y = 4\). Simplifying this, we find \(y = 4\). Thus, the y-intercept is \( (0, 4) \).
02

Find the x-intercepts

To find the x-intercepts, set \(y = 0\). The equation becomes \(4x^2 + 0 = 4\), which simplifies to \(4x^2 = 4\). Dividing both sides by 4, we get \(x^2 = 1\). Taking the square root of both sides, \(x = \pm1\). Thus, the x-intercepts are \( (1, 0) \) and \(( -1, 0 )\).
03

Plot the intercepts

On a coordinate plane, plot the points found: \( (0, 4) \), \((1, 0) \), and \(( -1, 0)\). These are the intercepts where the graph intersects the axes.
04

Plot additional points and draw the curve

Choose additional values for \(x\) (e.g., \( -2 \), \( 2 \)) and calculate corresponding \(y\) values to get a more accurate graph. For \(x = 2\), \(4(2^2) + y = 4\) simplifies to \(16 + y = 4\), so \(y = -12\). For \(x = -2\), the calculations are the same and give \(y = -12\). Now plot the points \( (2, -12) \) and \((-2, -12) \). Connect all plotted points to reveal the parabola.
05

Label the intercepts

Label the intercept points on the graph: \((0, 4)\), \((1, 0)\), and \((-1, 0)\). This clearly identifies where the graph intersects the axes.

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Key Concepts

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

Intercepts
Understanding intercepts is essential for graphing any function, including parabolas. Intercepts are the points where the graph crosses the axes. The **y-intercept** is where the graph intersects the y-axis. You can find it by setting the value of x to 0 in the equation.
  • For example, in the equation \(4x^2 + y = 4\), setting \(x = 0\) simplifies to \(4(0)^2 + y = 4\), giving us \(y = 4\). So, the y-intercept is \((0, 4)\).
The **x-intercepts** are where the graph intersects the x-axis. You can find them by setting the value of y to 0 in the equation.
  • For the same equation, setting \(y = 0\) simplifies to \(4x^2 + 0 = 4\), which further simplifies to \(4x^2 = 4\). Solving for x, we get \(x = \pm 1\). Thus, the x-intercepts are \((1, 0)\) and \((-1, 0)\).
Labeling these intercepts on the graph helps to clearly identify where the graph intersects the axes.
Quadratic Equations
Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). When graphing, these equations form parabolas.
In the example \(4x^2 + y = 4\), you can rewrite this as \(4x^2 + y - 4 = 0\), which is a standard quadratic equation without the linear term (bx).
  • The parabola opens upwards if the coefficient \(a\) is positive.
  • If \(a\) were negative, the parabola would open downwards.
Understanding the direction in which the parabola opens is crucial for plotting the points accurately. Knowing this can provide you with a heads-up on whether your found points make sense and align with the equation's properties.
Plotting Points
Plotting points is a crucial step in graphing any function, including parabolas. This step gives you a visual representation of the equation. After finding the intercepts, choose additional x-values to find more points on the graph.
  • For example, using the equation \(4x^2 + y = 4\), we pick \(x = 2\). The equation becomes, \(4(2^2) + y = 4\), simplifying to \(16 + y = 4\), so \(y = -12\).
  • Do the same for \(x = -2\), and you'll also get \(y = -12\).
Now plot these points \((2, -12)\) and \((-2, -12)\) on the graph.
Connect the plotted points to form a smooth curve, revealing the parabola. This method ensures you have an accurate representation, which includes the intercepts and any additional points necessary to complete the curve.
Remember to label all the points you've plotted, especially the intercepts, to clearly show where the graph intersects the axes.

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