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Magazine Chapter 11: Problem 22
Sketch the graph of the given equation. Find the intercepts; approximate to the nearest tenth where necessary. $$y=-x^{2}-2 x+3$$
Short Answer
Expert verified
y-intercept: (0, 3); x-intercepts: (-3, 0) and (1, 0). Vertex: (1, 0).
Step by step solution
01
Identify the general form
The given equation is in the form of a quadratic equation: \( y = -x^2 - 2x + 3 \).
02
Find the y-intercept
For any function, the y-intercept is the value of y when x = 0. Substitute x = 0 into the equation: \( y = -0^2 - 2(0) + 3 = 3 \). Therefore, the y-intercept is (0, 3).
03
Find the x-intercepts
The x-intercepts are the points where y = 0. Set the equation to 0 and solve for x: \( 0 = -x^2 - 2x + 3 \). This is a quadratic equation and can be solved using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -1\), \(b = -2\), and \(c = 3\). Plugging in the values, we get: \(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-1)(3)}}{2(-1)}\) \(x = \frac{2 \pm \sqrt{4 + 12}}{-2}\) \(x = \frac{2 \pm \sqrt{16}}{-2}\) \(x = \frac{2 \pm 4}{-2}\) So, \(x = \frac{2 + 4}{-2} = -3\) and \(x = \frac{2 - 4}{-2} = 1\). Therefore, the x-intercepts are (-3, 0) and (1, 0).
04
Find the vertex
The vertex form of a quadratic equation \(y = ax^2 + bx + c\) can be used to find the vertex, with the x-coordinate of the vertex given by \(x = -\frac{b}{2a}\). For our equation, \(a = -1\) and \(b = -2\): \(x = -\frac{-2}{2(-1)} = 1\). Substitute x = 1 back into the original equation to find the y-coordinate: \(y = -1^2 - 2(1) + 3 = -1 - 2 + 3 = 0\). Therefore, the vertex is (1, 0).
05
Sketch the graph
Now that the intercepts and vertex are known, plot the points (0, 3), (-3, 0), (1, 0), and the vertex (1, 0) on a graph. Draw a smooth parabolic curve passing through these points, opening downwards because the coefficient of \(x^2\) is negative.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
y-intercept
The y-intercept of a quadratic equation is the point where the graph crosses the y-axis. To find this point, we set x to 0 and solve for y.
For the equation given, which is \( y = -x^2 - 2x + 3 \), substitute x = 0. This simplifies to \( y = 3 \).
This tells us that when x is 0, y is 3. So, the y-intercept is at the point (0, 3).
Keep in mind, finding the y-intercept is always straightforward for any quadratic equation: simply set x to 0 and solve the equation.
For the equation given, which is \( y = -x^2 - 2x + 3 \), substitute x = 0. This simplifies to \( y = 3 \).
This tells us that when x is 0, y is 3. So, the y-intercept is at the point (0, 3).
Keep in mind, finding the y-intercept is always straightforward for any quadratic equation: simply set x to 0 and solve the equation.
x-intercepts
The x-intercepts are the points where the graph crosses the x-axis. This happens where y is 0.
To find the x-intercepts for \( y = -x^2 - 2x + 3 \), set y to 0 and solve for x.
So, we need to solve \( 0 = -x^2 - 2x + 3 \). This is a quadratic equation, and we solve it using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation, a = -1, b = -2, and c = 3. Plugging these values into the formula, we get:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-1)(3)}}{2(-1)} \]
This simplifies to:
\[ x = \frac{2 \pm \sqrt{4 + 12}}{-2} \]
Then, \[ x = \frac{2 \pm \sqrt{16}}{-2} \]
Finally, \[ x = \frac{2 \pm 4}{-2} \]
This gives us two solutions: \[ x = \frac{2 + 4}{-2} = -3 \] and \[ x = \frac{2 - 4}{-2} = 1 \].
Thus, the x-intercepts are at (-3, 0) and (1, 0).
To find the x-intercepts for \( y = -x^2 - 2x + 3 \), set y to 0 and solve for x.
So, we need to solve \( 0 = -x^2 - 2x + 3 \). This is a quadratic equation, and we solve it using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation, a = -1, b = -2, and c = 3. Plugging these values into the formula, we get:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-1)(3)}}{2(-1)} \]
This simplifies to:
\[ x = \frac{2 \pm \sqrt{4 + 12}}{-2} \]
Then, \[ x = \frac{2 \pm \sqrt{16}}{-2} \]
Finally, \[ x = \frac{2 \pm 4}{-2} \]
This gives us two solutions: \[ x = \frac{2 + 4}{-2} = -3 \] and \[ x = \frac{2 - 4}{-2} = 1 \].
Thus, the x-intercepts are at (-3, 0) and (1, 0).
vertex
The vertex of a parabola is its highest or lowest point, depending on whether it opens upwards or downwards.
For the standard quadratic equation, the vertex can be found using the formula: \[ x = \frac{-b}{2a} \]
For our equation \( y = -x^2 - 2x + 3 \), a is -1 and b is -2.
Plugging these into the formula, we get: \[ x = \frac{-(-2)}{2(-1)} = \frac{2}{-2} = -1 \]
This tells us the x-coordinate of the vertex is -1.
To find the y-coordinate of the vertex, substitute x = -1 back into the equation: \[ y = -(-1)^2 - 2(-1) + 3 = -1 + 2 + 3 = 4 \]
Thus, the vertex is at (-1, 4).
The vertex is a crucial point as it indicates the maximum or minimum value of the quadratic function.
For the standard quadratic equation, the vertex can be found using the formula: \[ x = \frac{-b}{2a} \]
For our equation \( y = -x^2 - 2x + 3 \), a is -1 and b is -2.
Plugging these into the formula, we get: \[ x = \frac{-(-2)}{2(-1)} = \frac{2}{-2} = -1 \]
This tells us the x-coordinate of the vertex is -1.
To find the y-coordinate of the vertex, substitute x = -1 back into the equation: \[ y = -(-1)^2 - 2(-1) + 3 = -1 + 2 + 3 = 4 \]
Thus, the vertex is at (-1, 4).
The vertex is a crucial point as it indicates the maximum or minimum value of the quadratic function.
parabola
A parabola is the graph of a quadratic function. It is a U-shaped curve that can open upwards or downwards.
For our equation \( y = -x^2 - 2x + 3 \), the parabola opens downwards because the coefficient of \( x^2 \) is negative.
Key characteristics of a parabola include its vertex, y-intercept, and x-intercepts.
These points help in sketching the graph accurately.
In our case, the important points are the y-intercept (0, 3), x-intercepts (-3, 0) and (1, 0), and vertex (-1, 4).
Plot these points and draw a smooth curve through them to create the parabola.
The points where the parabola intersects the axes are helpful in understanding the graph's shape and position.
For our equation \( y = -x^2 - 2x + 3 \), the parabola opens downwards because the coefficient of \( x^2 \) is negative.
Key characteristics of a parabola include its vertex, y-intercept, and x-intercepts.
These points help in sketching the graph accurately.
In our case, the important points are the y-intercept (0, 3), x-intercepts (-3, 0) and (1, 0), and vertex (-1, 4).
Plot these points and draw a smooth curve through them to create the parabola.
The points where the parabola intersects the axes are helpful in understanding the graph's shape and position.
