Question

How do you graph the parabola $$f\left(x\right)=x^2$$?

Answer 1

Hint: In this given problem we have to graph the given parabola. The given equation of the parabola $$y=x^2$$ has the direction open upwards so we can draw an open upwards curve in the graph with its vertex at the origin as the x-intercept and y-intercept of this equation is 0. We can also know that if x is 0 y is also 0. Therefore, the vertex must be the origin. We assume values for x to find the value of y and plot the graph.

Complete step by step answer:
We know that the given equation of the parabola is,
$$y=x^2$$ …….. (1)
We also know that when x is equal to 0 y is also equal to 0 in the above equation, the vertex is the maximum/minimum point of the curve, therefore the vertex is at the origin $$(0,0)$$.
We can assume a value for x to get the point to be plotted in the graph.
When x = 1, the value of y from the equation (1) is,
$$\Rightarrow y=1$$
The point is $$(1,1)$$.
When x = 2, the value of y from the equation (1) is,
$$\begin{array}{rl}& \Rightarrow y=2^2\\ & \Rightarrow y=4\end{array}$$
The point is $$(2,4)$$.
When x = -1, the value of y from the equation (1) is,
$$\begin{array}{rl}& \Rightarrow y={(-1)}^2\\ & \Rightarrow y=1\end{array}$$
The point is $$(-1,1)$$.
When x = -2, the value of y from the equation (1) is,
$$\begin{array}{rl}& \Rightarrow y={(-2)}^2\\ & \Rightarrow y=4\end{array}$$
The point is $$(-2,4)$$.
Therefore, the points $$(1,1)$$, $$(-1,1)$$, $$(-2,4)$$ ,$$(2,4)$$ can be plotted in the $$y=x^2$$graph.

Note:
 Students make mistakes while assuming the values for x to solve for y, which should be concentrated. Students may also make mistakes while plotting the points in the graph. We should check for the vertex point first to plot the curve. We should plot the graph with the correct values.