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Chapter 3: Problem 49

Identify the vertex, axis of symmetry, y-intercept, x-intercepts, and opening of each parabola, then sketch the graph. $$y=-3(x-2)^{2}+12$$

Short Answer

Expert verified
Vertex: (2, 12); Axis of symmetry: x = 2; Y-intercept: (0, 0); No x-intercepts; Opens downwards.

Step by step solution

01

- Identify the Vertex

The general form of a parabola equation in vertex form is \[ y = a(x-h)^2 + k \]In the given equation, \[ y = -3(x-2)^{2}+12 \]Comparing this with the general form, you can extract the vertex (h, k) as (2, 12).
02

- Identify the Axis of Symmetry

The axis of symmetry of a parabola in vertex form \[ y=a(x-h)^2+k \]is given by the line x = h. For the given equation, the axis of symmetry is \[ x = 2 \].
03

- Identify the Y-Intercept

To find the y-intercept, set x to 0 and solve for y:\[ y=-3(0-2)^{2}+12 \]Solving this gives: \[ y = -3(4) + 12 = -12 + 12 = 0 \]Thus, the y-intercept is (0, 0).
04

- Identify the X-Intercepts

To find the x-intercepts, set y to 0 and solve for x:\[ 0 = -3(x-2)^2 + 12 \]Isolate the squared term:\[ 12 = 3(x-2)^2 \]Divide both sides by -3:\[ -4 = (x-2)^2 \]Since squaring a real number cannot yield a negative value, there are no real x-intercepts.
05

- Determine the Opening Direction

The coefficient a in the equation \[ y = a(x-h)^2 + k \]indicates the direction of the parabola's opening. For \[ y = -3(x-2)^2 + 12 \]since \[ a = -3 \], which is negative, the parabola opens downwards.
06

- Sketch the Graph

Using the identified vertex (2, 12), axis of symmetry x = 2, y-intercept (0, 0), and the fact that the parabola opens downwards, sketch the graph. Plot the vertex, draw the axis of symmetry, and sketch the parabola opening downwards without x-intercepts.

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Key Concepts

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

vertex of parabola
The vertex of a parabola represents its highest or lowest point. It is provided in the form \( (h, k) \) in the vertex form equation \[ y = a(x - h)^2 + k \]. For the given equation \[ y = -3(x-2)^2 + 12 \], you can easily find the vertex by identifying \ h \ and \ k \. Here, the vertex is at (2, 12). This means the highest point of the parabola (since a < 0) is at (2, 12).
axis of symmetry
The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. This line passes through the vertex and is expressed as \[ x = h \] where \ h \ is the x-coordinate of the vertex. In our case, since the vertex is (2, 12), the axis of symmetry is \[ x = 2 \]. This line helps in sketching the parabola accurately.
y-intercept calculation
The y-intercept is the point where the parabola crosses the y-axis. To find it, set \ x = 0 \ and solve for \ y \. For the equation \[ y = -3(x-2)^2+12 \], substitute \ x = 0 \ into the equation: \[ y = -3(0-2)^2+12 \]. Simplifying, we get \[ y = -3(4) + 12 = -12 + 12 = 0 \]. Thus, the y-intercept is \(0, 0 \).
x-intercepts
To find the x-intercepts, set \ y \ to 0 and solve for \ x \. For the equation \[ 0 = -3(x-2)^2+12 \], isolate the squared term: \[ 12 = 3(x-2)^2 \]. Dividing both sides by -3 gives \[ -4 = (x-2)^2 \]. Since you can't square a number and get a negative result, this equation has no real solutions. Thus, there are no real x-intercepts for this parabola.
parabola opening direction
The direction in which a parabola opens is determined by the coefficient \ a \ in the vertex form of the equation \ y = a(x-h)^2 + k \. If \ a \ is positive, the parabola opens upwards; if \ a \ is negative, it opens downwards. In our example \[ y = -3(x-2)^2+12 \], \ a = -3 \, which is negative. Thus, the parabola opens downwards. This tells us that the vertex at (2, 12) is the highest point on the curve.

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