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Question: How do you graph the parabola \[y = - \dfrac{1}{2}{x^2}\] using vertex, intercepts and additional po...

How do you graph the parabola y=−12x2y = - \dfrac{1}{2}{x^2} using vertex, intercepts and additional points?

Explanation

Solution

Here in this question, we have to determine the vertex and intercepts for the given equation. By substituting the x is equal to zero we can find the y intercept value and substituting y is equal to zero we can find the x intercept value. By equating the given equation to the general vertex equation we can determine the vertex.

Complete step by step answer:
The given equation is in the form of f(x)=a(x−h)2+kf(x) = a{(x - h)^2} + k, where (h, k) represents the vertex for the equation. An intercept is a point where the straight line or a curve intersects the y-axis in a plane. If the point x is zero then the obtained point is a y -intercept.
Now consider the given equation y=−12x2y = - \dfrac{1}{2}{x^2} -----------(1)
Substitute the value of x as 0 in equation (1), then we have
⇒y=−12(0)2\Rightarrow y = - \dfrac{1}{2}{(0)^2}
On simplifying we get
⇒y=0\Rightarrow y = 0
Therefore, y-intercept is (0,0)
Substitute the value of y as 0 in the equation (1) then we have
⇒0=−12x2\Rightarrow 0 = - \dfrac{1}{2}{x^2}
On simplifying we get
⇒x=0\Rightarrow x = 0
Therefore, x-intercept is (0, 0)
The general vertex equation of a line is given byf(x)=a(x−h)2+kf(x) = a{(x - h)^2} + k,----- (2) where (h, k) is a vertex
Since the given equation of a graph is a not having the constant term the vertex will be
(h,k)=(0,0)(h,k) = (0,0)
Therefore, the vertex is (0,0)(0,0)
we find the value of y by using the graph equation y=−12x2y = - \dfrac{1}{2}{x^2}. Let we substitute the value of x has -2, -1, 0, 1 and 2
Now we consider the value of x as -2, the value of y is

⇒y=−12(−2)2 ⇒y=−2  \Rightarrow y = - \dfrac{1}{2}{( - 2)^2} \\\ \Rightarrow y = - 2 \\\

Now we consider the value of x as -1, the value of y is

⇒y=−12(1)2 ⇒y=−0.5  \Rightarrow y = - \dfrac{1}{2}{(1)^2} \\\ \Rightarrow y = - 0.5 \\\

Now we consider the value of x as 0, the value of y is

⇒y=−12(0)2 ⇒y=0  \Rightarrow y = - \dfrac{1}{2}{(0)^2} \\\ \Rightarrow y = 0 \\\

Now we consider the value of x as 1, the value of y is

⇒y=−12(1)2 ⇒y=−0.5  \Rightarrow y = - \dfrac{1}{2}{(1)^2} \\\ \Rightarrow y = - 0.5 \\\

Now we consider the value of x as 1, the value of y is

⇒y=−12(2)2 ⇒y=−2  \Rightarrow y = - \dfrac{1}{2}{(2)^2} \\\ \Rightarrow y = - 2 \\\

Now we draw a table for these values we have

X-2-1012
y-1-0.50-0.5-1

The graph plotted for this point is represented below:

Note: The graph is plotted x-axis versus y axis. The graph is two dimensional. By the equation of a graph, we can plot the graph by assuming the value of x. We can’t assume the value of y. because the value of y depends on the value of x. hence, we have plotted the graph.