Question
Question: What is the nature of the graph: \(y = - 4{x^2} + 6\) A. Parabola not passing through origin B. ...
What is the nature of the graph: y=−4x2+6
A. Parabola not passing through origin
B. Hyperbola not passing through origin
C. Ellipse not passing through origin
D. It is not a conic.
Solution
To find the nature of the graph y=−4x2+6, let us first draw the graph for different values of x. From the graph we can clearly observe the nature of y=−4x2+6. The given function will be a parabola if it is mirror-symmetrical with a single open curve that extends till infinity and is U-shaped. The function will be a hyperbola if it is an open curve with two unconnected branches and has two axes of symmetry. y=−4x2+6 will be an ellipse if it is a planar curve that has two focal points and appears to be somewhat like a circle.
Complete step by step answer:
We need to find the nature of the graph y=−4x2+6. Let us first draw this graph. For this, first let us find the coordinates to plot the graph. The table below shows the values of y=−4x2+6 for different values of x.
x | y=−4x2+6 |
---|---|
−1 | 2 |
0 | 6 |
1 | 2 |
From the graph, we can infer that the shape is a U-shape and it extends to infinity. It does not have two axes of symmetry or like a circle. Hence, the given function is a parabola. Now, let us see whether it passes through the origin or not. We know that at the origin x and y coordinates are 0. That is, the point (0,0). From the graph, we can see that the parabola passes through the point (0,6) not through the origin.Hence, y=−4x2+6 is a parabola not passing through the origin.
Therefore, the correct option is A.
Note: Do not get confused with parabola and hyperbola. A hyperbola will have the equation of the form a2x2−b2y2=1. The graph of this will be like:
Also, remember that when either x or y is squared but not both, the given equation will be an equation of parabola. When x and y are both squared and the coefficients are positive but different, then the given equation is the equation of ellipse. When x and y are both squared, and exactly one of the coefficients is negative and exactly one of the coefficients is positive, then it resembles a hyperbola.