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Chapter 11: Problem 3

Sketch the graph of the given equation. Find the intercepts; approximate to the nearest tenth where necessary. $$y=x^{2}+2$$

Short Answer

Expert verified
The y-intercept is (0, 2). There are no real x-intercepts. The parabola opens upwards with vertex at (0, 2).

Step by step solution

01

Identify the type of graph

The given equation is a quadratic equation in the form of y = x^2 + 2This represents a parabola.
02

Find the y-intercept

To find the y-intercept, set x = 0 and solve for y. Substitute x = 0 into the equation: y = (0)^2 + 2 = 2Thus, the y-intercept is (0, 2).
03

Find the x-intercepts

To find the x-intercepts, set y = 0 and solve for x. 0 = x^2 + 2We solve for x and get:x^2 = -2Since the square of a real number cannot be negative, the equation has no real x-intercepts.
04

Graph the parabola

The vertex of the parabola is the minimum point, occurring at (0, 2). Since the coefficient of x^2 is positive, the parabola opens upwards. Plot the vertex at (0, 2) and sketch the parabola opening upwards.

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Key Concepts

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

Parabola
A parabola is a U-shaped curve that can open either upwards or downwards. In the context of quadratic equations like in our exercise, the equation generally takes the form of \(y = ax^2 + bx + c\). Here, \(a\) determines the direction of the parabola:
  • If \(a > 0\), the parabola opens upwards.
  • If \(a < 0\), the parabola opens downwards.
In our exercise, the equation \(y = x^2 + 2\) has \(a = 1\) (positive), so the parabola opens upwards.
Y-Intercept
The y-intercept is the point where the graph crosses the y-axis. To find it, we set \(x = 0\) and solve for \(y\). This is because the y-axis occurs where \(x = 0\). In our equation \(y = x^2 + 2\), substituting \(x = 0\) gives us:
\[y = (0)^2 + 2 = 2 \]
Thus the y-intercept is at (0, 2). This is a critical point to plot on the graph for sketching the parabola.
X-Intercept
The x-intercept is where the graph crosses the x-axis. To find the x-intercept, we set \(y = 0\) and solve for \(x\). Setting the given equation to zero, we have: \[0 = x^2 + 2 \]
Solving for x, we find:
\[x^2 = -2 \]
Since the square of any real number cannot be negative, this equation has no real solutions. Thus, the equation \(y = x^2 + 2\) has no x-intercepts. This indicates that the parabola does not cross the x-axis, which is important when sketching the graph.
Vertex
The vertex is the high or low point of the parabola, located at \( (h,k) \) in the general vertex form \( y = a(x - h)^2 + k \). In our exercise, the given equation \(y = x^2 + 2\) is already in the simplified form \(y = (x - 0)^2 + 2\). Hence, the vertex is at \( (0,2) \).
  • This suggests that the minimum point of the parabola is at (0,2).
  • The coefficient of \(x^2\) is positive, thus the parabola opens upwards.
Plotting the vertex gives us a key reference point for sketching the parabola.

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Most popular questions from this chapter

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