Problem 76 Find the vertex, axis of symmetr... [FREE SOLUTION]

Chapter 10: Problem 76

Find the vertex, axis of symmetry, $x$ -intercept, $y$ -intercepts, focus, and directrix for each parabola. Sketch the graph, showing the focus and directrix. $$x=-\frac{1}{4}(y-2)^{2}-1$$

Short Answer

Expert verified

Vertex: $(-1,2)$,Axis of Symmetry: $y = 2$, x-intercept: $(-2,0)$, no y-intercept. Focus: $(-2,2)$ Directrix: $x = 0$.

Step by step solution

Identify the vertex

The given equation is already in the vertex form for a parabola that opens to the left or right: i.e., $x = a(y鈭択)^2 + h$By comparing the given equation $x = -\frac{1}{4}(y - 2)^2 - 1$,with this form, you can identify the vertex at $(h,k) = (-1,2)$.

Determine the axis of symmetry

The axis of symmetry for a parabola that opens left or right is given by the equation $y = k$. Here, with vertex $(h,k) = (-1,2)$, the axis of symmetry is $y = 2$.

Find the x-intercept

To find the x-intercept, set $y = 0$ and solve for $x$: \[ x = -\frac{1}{4}(0-2)^2 - 1 \] \[ x = -1 -1 \] \[x = -2\].So, the x-intercept is $(-2, 0)$.

Find the y-intercept

To find the y-intercept, set $x = 0$ and solve for $y$: \[ 0 = -\frac{1}{4}(y-2)^2 - 1 \] \[ 1 = -\frac{1}{4}(y-2)^2 \] \[(y-2)^2 = -4\].Since the square of a real number cannot be negative,there is no real solution here.Thus, there is no y-intercept.

Find the focus and directrix

The standard form for a horizontally oriented parabola is: \[ x-h = -\frac{1}{4}(y-k)^2 \] given $ x = -\frac{1}{4}(y-2)^2 - 1 $.Here, $a = -\frac{1}{4}$ - comparing with \[x - h = a (y - k)^2 \], we notice that $4a = -1$ thus, $4a = -4p \Rightarrow p = 1$.Since $ a = -1/(4p) \Rightarrow 1/4 = p$ - which means the parabola opens to the left with a value of $[-1,2]$.

Sketch the graph by Showing Focus and Directrix

The vertex is $(-1, 2)$, the focus is $(-2, 2)$, and the directrix is $x = 0$.To sketch:1. Plot the vertex at $(-1, 2)$.2. Plot the focus at $(-2,2)$.3. Draw the directrix line at $x = 0$.4. Draw a parabola opening to the left around the vertex, focusing through the vertex.

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Key Concepts

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

vertex form of parabola

The vertex form of a parabola equation is a special format that makes it easy to identify key features. It looks like this: $ x = a(y鈭択)^2 + h $ or $ y = a(x鈭抙)^2 + k $. Here, $ (h, k) $ are the coordinates of the vertex of the parabola. For example, in the equation $ x = -\frac{1}{4}(y - 2)^2 - 1 $, comparing this with the general form, we find the vertex is $ (h, k) = (-1, 2) $. The coefficient $ a $ indicates the direction in which the parabola opens. If $ a $ is positive, it opens upward or to the right. If negative, it opens downward or to the left.

axis of symmetry

The axis of symmetry is a vertical or horizontal line that passes through the vertex of the parabola, dividing it into two symmetrical halves. For parabolas in the form $ x = a(y鈭択)^2 + h $, the axis of symmetry is given by $ y = k $. For our example equation $ x = -\frac{1}{4}(y - 2)^2 - 1 $, the axis of symmetry is $ y = 2 $. This means the parabola is symmetric about the line $ y = 2 $.

parabola intercepts

Intercepts are the points where the parabola intersects the x-axis and y-axis.

To find the x-intercept, set $ y = 0 $ in the equation. For $ x = -\frac{1}{4}(0-2)^2 - 1 $, solving gives $ x = -2 $. So, the x-intercept is $ (-2, 0) $.
To find the y-intercept, set $ x = 0 $. For $ 0 = -\frac{1}{4}(y-2)^2 - 1 $, we solve and see there's no real solution for $ y $. Therefore, the parabola does not have a y-intercept.

parabola focus and directrix

The focus and directrix are unique features of parabolas that help define their shape. The focus is a point, and the directrix is a line perpendicular to the axis of symmetry.

For our equation $ x = -\frac{1}{4}(y - 2)^2 - 1 $, we found that $ a = -\frac{1}{4} $, and comparing with the standard form, $ 4p = 1 $, which means $ p = 1/4 $. The focus is then determined by moving 1 unit left (since our parabola opens left) from the vertex, giving us $ (-2, 2) $.
The directrix is a vertical line equivalent to moving 1 unit right (opposite the direction of the focus) from the vertex, giving us $ x = 0 $.

graphing parabolas

Graphing a parabola involves plotting its key features and understanding its shape. Here's how you can graph the given equation:

Start by plotting the vertex at $ (-1, 2) $.
Next, plot the focus at $ (-2, 2) $ and draw the directrix line $ x = 0 $.
Draw the axis of symmetry line $ y = 2 $.
Using the vertex and the direction the parabola opens (leftward in this case), sketch the curve of the parabola so it passes through the vertex and around the focus.
Add the x-intercept point $ (-2, 0) $ to complete the graph.

91影视

Find the vertex, axis of symmetry, \(x\) -intercept, \(y\) -intercepts, focus, and directrix for each parabola. Sketch the graph, showing the focus and directrix. $$x=-\frac{1}{4}(y-2)^{2}-1$$

Short Answer

Step by step solution

Identify the vertex

Determine the axis of symmetry

Find the x-intercept

Find the y-intercept

Find the focus and directrix

Sketch the graph by Showing Focus and Directrix

Key Concepts

vertex form of parabola

axis of symmetry

parabola intercepts

parabola focus and directrix

graphing parabolas

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