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Chapter 0: Problem 7

Sketch the graphs of the equations. $$ 2 y-x^{2}=1 $$

Short Answer

Expert verified
The graph of the given equation \(y = \frac{1}{2}x^2 + \frac{1}{2}\) is a parabola opening upwards. The y-intercept is at the point (0, 1/2), the x-intercepts are at the points (-1, 0) and (1, 0), the axis of symmetry is the vertical line x = 0, and the vertex is at the point (0, 1/2).

Step by step solution

01

Rewrite the equation in terms of y

First, we need to rewrite the given equation in terms of y for easier visualization. The given equation is: \(2y - x^2 = 1\) Adding \(x^2\) to both sides, we have: \(2y = x^2 + 1\) Now, dividing both sides by 2: \(y = \frac{1}{2}x^2 + \frac{1}{2}\)
02

Determine the y- and x-intercepts

To find the y-intercept, we set x to 0 in the equation and solve for y. \(y = \frac{1}{2}(0)^2 + \frac{1}{2}\) So, the y-intercept is on the point (0, 1/2). To find the x-intercepts, set y to 0 in the equation. \(0 = \frac{1}{2}x^2 + \frac{1}{2}\) Subtracting \(\frac{1}{2}\) from both sides: \(-\frac{1}{2} = \frac{1}{2}x^2\) Now, multiplying both sides by -2 to eliminate the fractions: \(1 = x^2\) Taking the square root of both sides: \(x = \pm \sqrt{1}\) Thus, we have two x-intercepts, at the points (-1, 0) and (1, 0).
03

Determine the axis of symmetry and vertex

Since this is a quadratic equation in the form \(y = ax^2 + bx + c\), and the coefficient of the x^2 term is positive, we know that the parabola opens upwards. The axis of symmetry (AoS) for a quadratic equation is given by \(x = \frac{-b}{2a}\), with a and b being the coefficients of the x^2 and x terms, respectively. In our case, a = (1/2) and b = 0. Thus, the AoS is x = 0. The vertex of the parabola can be found using the equation \(y = a(x - h)^2 + k\), where (h, k) represents the vertex and a is the coefficient of the x^2 term. For our current equation, we have: \(y = \frac{1}{2}(x - 0)^2 + 0.5\) So, the vertex of the parabola is at the point (0, 1/2).
04

Sketch the graph

Now that we have gathered all the necessary information, we can sketch the graph of the given equation: 1. Plot the y-intercept point (0, 1/2) 2. Plot the x-intercept points (-1, 0) and (1, 0) 3. Draw the axis of symmetry (a vertical line at x = 0) 4. Plot the vertex (0, 1/2) 5. Draw the parabola using the x-axis intercepts and vertex. Our final sketch of the graph should show that it is a parabola opening upwards, and the vertex is at the point (0, 1/2).

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