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Chapter 0: Problem 43

Sketch the graph of the equation. Identify any intercepts and test for symmetry. $$y=1-x^{2}$$

Short Answer

Expert verified
The graph is a parabola opening downwards, symmetric with respect to the y-axis, and passing through the points (-1,0), (0,1), and (1,0). These points are respectively the x-intercepts and y-intercept for the equation \(y=1-x^{2}\).

Step by step solution

01

Identify Y-Intercept

Substitute \(x=0\) into the equation, which will yield the y-intercept. So, if in the equation \(y=1-x^{2}\), \(x\) is replaced by \(0\), then \(y\) equals \(1\). So, the y-intercept is \(1\).
02

Identify X-Intercept(s)

To find x-intercept(s), replace \(y\) with \(0\) in the equation.\(0 = 1 - x^{2}\) \nSolving for \(x\), we get \(x = \pm \sqrt{1} = \pm 1\). Thus, the x-intercepts are \(\-1\) and \(1\).
03

Plot the Intercept Points

Plot the y-intercept point, which is at \((0,1)\) on the graph. Also, plot the x-intercepts, which are at \((-1, 0)\) and \((1, 0)\). This will give two points on the graph and should assist in the sketching of the graph.
04

Test for Symmetry

To determine the symmetry, first replace \(x\) with \(-x\) in the equation \(y=1-x^2\), which will yield \(y=1-(-x)^2\). Simplifying this yields the original equation, indicating the graph is symmetric about the y-axis. This will assist in generating an accurate sketch of the graph.
05

Sketch the Graph

Now that we know the graph is symmetric about the y-axis, we can draw a smooth curve passing through the points \((-1,0)\), \((0,1)\), and \((1,0)\). This will be a parabola opening downwards as the coefficient of \(x^2\) is negative. With y-intercept as the highest point, the curve should extend down and away on both sides symmetrically determining 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.

Intercepts
When understanding quadratic graphs, one crucial aspect to grasp is the concept of intercepts. Intercepts are points where the graph touches the axes, meaning the x-axis and the y-axis.
  • Y-intercept: To find where the graph crosses the y-axis, you set the value of x to 0. This helps us see what y equals when x is zero. In the equation \(y = 1 - x^2\), substituting \(x = 0\) gives \(y = 1\). Thus, the y-intercept is at the point \((0, 1)\).
  • X-intercepts: On the flip side, to find the x-intercepts where the graph crosses the x-axis, we set y to 0. So, you solve \(0 = 1 - x^2\). Solving for x gives us \(x = \pm 1\). Thus, the x-intercepts are the points \((-1, 0)\) and \((1, 0)\).

Identifying these intercepts provides not just points of contact with the axes, but also critical points to begin sketching and understanding the graph overall.
Symmetry
Symmetry is another key feature of quadratic graphs, especially parabolas. Symmetry generally makes these graphs easier to understand and plot.
  • A graph is symmetric about the y-axis if replacing \(x\) with \(-x\) still leads to the same equation. This means whatever is on one side of the y-axis will be mirrored exactly on the other side.

In our equation \(y = 1 - x^2\), replace \(x\) with \(-x\). You will then have \(y = 1 - (-x)^2\), which simplifies back to \(y = 1 - x^2\).

Since it's unchanged, it indicates that this parabola is symmetric about the y-axis. This symmetry helps us draw the graph correctly by knowing if we chart points on one side, we can mirror them on the other.
Parabola Sketching
Sketching parabolas becomes straightforward once you understand the curve's behavior and intercepts. In the case of \(y = 1 - x^2\), the following steps make it easier:
  • Begin by plotting the intercepts: \((0, 1)\), \((-1, 0)\), and \((1, 0)\).
  • Recognize that the parabola is dictated by the negative coefficient of \(x^2\), meaning it opens downwards.
  • Given that the graph is symmetric about the y-axis, ensure that whatever points you plot on one side are mirrored precisely on the other side.

With these steps, complete the sketch by drawing a smooth curve that connects these points. The y-intercept, being the highest point, will assist in ensuring that it peaks at \((0, 1)\) before swooping down symmetrically through the x-intercepts.

Parabola sketching becomes a fun exercise in symmetry and understanding the equation's basic properties!

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Most popular questions from this chapter

Apartment Rental A real estate office handles an apartment complex with 50 units. When the rent is $$\$ 580$$ per month, all 50 units are occupied. However, when the rent is $$\$ 625$$, the average number of occupied units drops to 47 . Assume that the relationship between the monthly rent \(p\) and the demand \(x\) is linear. (Note: The term demand refers to the number of occupied units.) (a) Write a linear equation giving the demand \(x\) in terms of the rent \(p\). (b) Linear extrapolation Use a graphing utility to graph the demand equation and use the trace feature to predict the number of units occupied if the rent is raised to $$\$ 655$$. (c) Linear interpolation Predict the number of units occupied if the rent is lowered to $$\$ 595$$. Verify graphically.

Evaluate (if possible) the function at the given value(s) of the independent variable. Simplify the results. \(g(x)=3-x^{2}\) (a) \(g(0)\) (b) \(g(\sqrt{3})\) (c) \(g(-2)\) (d) \(g(t-1)\)

Determine whether \(y\) is a function of \(x\). $$x^{2}+y=4$$

Determine whether \(y\) is a function of \(x\). $$x^{2}+y^{2}=4$$

Sketch a graph of the function and find its domain and range. Use a graphing utility to verify your graph. $$f(x)=\sqrt{9-x^{2}}$$
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