দ্বিঘাত সমীকরণের গ্রাফ:$y=ax^2$

All quadratic functions have the form

$$y = ax^2 + bx + c,$$

where $a, b, c$ are real numbers. If any of them are zero, one of the terms goes away.

If $a = 0,$ then $y = bx + c.$ If $b = 0,$ then $y = ax^2 + c.$ If $c = 0,$ then $y = ax^2 + bx.$

The problem with $a = 0$ is that there is no longer a quadratic term. This makes the function linear.

We can make the most basic quadratic function by making $b$ and $c$ BOTH equal zero. This minimizes the number of terms. We are left with:

$$y = ax^2$$

Let's dive into this.

For parabolas, we care most about two things:

1. Where is the parabola's vertex?

2. Does the parabola point upward or downward?

We will talk about Question 2 first. The direction of the parabola only depends on whether $a$ is positive or negative.

If $a$ is positive, the parabola points upward. For example, consider this graph of $y = 2x^2$ ($a = 2$ here):

The graph suggests that $y = ax^2$ is never negative. We can demonstrate this with some algebra.

If $x = 0,$ then

$$y = a(0)^2 = a(0) = 0,$$

regardless of what $a$ is. If $x$ is nonzero, then $x^2$ is positive (try checking this for your favorite nonzero numbers). If $a$ is positive, then $y$ is just the product of two positive numbers, $a$ and $x^2$. This means $y$ is positive whenever $x \neq 0.$

We do almost the same thing if $a$ is negative rather than positive. Let's look at $y = -5x^2$ as an example (in this case $a = -5$).

Now we will explain why $y$ is never positive if $a$ is negative. If $x = 0$ then

$$y = ax^2 = a(0)^2 = a(0) = 0,$$

and now we are left with considering what happens if $x \neq 0.$ The value of $x^2$ is positive for any nonzero $x,$ so $y$ is a product of a negative number and a positive number. Therefore $y$ must be negative.

Now we're going to talk about the vertex. The vertex is the "high point" of the parabola if it points downwards, and the "low point" of the parabola if it points upwards.

Did you notice how $x = 0$ has "special" behavior on the parabola regardless of whether $a$ was positive or negative? Well, there's a reason for that. If $a$ is negative, then $y = ax^2$ is zero at $x = 0$ and negative everywhere else. Thus $y$ will be as big as possible if $x = 0,$ so this will be the vertex.

If $a$ is positive, then $y = ax^2$ is zero at $x = 0$ and positive everywhere else. Thus $y$ will be as small as possible if $x = 0,$ so this will be the vertex.

Here is a brief summary.

1. The "simplest" quadratic functions are those that look like $y = ax^2$.

2. If $a > 0,$ the parabola points upwards.

3. If $a < 0,$ the parabola points downwards.

4. $x = 0$ is the vertex of any parabola of the form $y = ax^2$.

The standard form of a quadratic function is the following:

$$y = ax^2+bx+c$$

$x$ represents an unknown variable, $a$, $b$, and $c$ are constants, and $a\ne 0$. When $b= 0$ and $c= 0$, the quadratic function is of the form

$$y = ax^2$$

The graph $y = ax^2$ takes the shape of a parabola. The sign of $a$ determines where the graph would be located. When $a>0$, the parabola is located above the $x$-axis. When $a<0$, the parabola is located below the $x$-axis. $y = ax^2$ and $y = -ax^2$ are symmetric about the $x$-axis.

Graphed below are the function $y = x^2$ (in blue) and $y = -x^2$ (in green).

The coefficient $a$ determines the concavity of the parabola (how wide or thin the parabola is). Concavity corresponds with the rate that a parabola opens.

When $\left|a\right| > 1$, the parabola is thinner because the function is increasing at a faster rate. For example, take the two functions $f(x) = 2x^2$ and $g(x) = x^2$. $f(1) > g(1)$, which shows that $f(x)$ increases at a higher rate.

When $\left|a\right| < 1$, the parabola is wider because the function is increasing at a slower rate (or decreasing at a faster rate). For example, take the two functions $f(x) = \frac{1}{2}x^2$ and $g(x) = x^2$. $f(1) < g(1)$, which shows that $f(x)$ increases at a lower rate.

Graphed below are the function $y = x^2$ in blue, $y = 3x^2$ in green, and $y = \frac{1}{3}x^2$ in red.

(Video) Algebra 2 4-1 Quadratic Functions and Transformations: Problem 1 - Graphing $f(x) = ax²$

by Matthew Richardson

This video by Matthew Richardson discusses the graph of $f(x)=ax^2$.

$\large{\text{Summary}}$

Every quadratic has the form $y=ax^2+bx+c,$ where $a, b,$ and $c$ are constants and $a \neq 0$ . Each coefficient in the equation gives us information about the graph of the parabola. There are values such as the vertex and axis of symmetry that need the $b$ and $c$ values. We're going to focus on information we can get when there is only the quadratic term, $ax^2$.

Matthew looks in detail at the function $f(x)=\frac12 x^2$. We can start by plugging numbers into the function and finding the output.

$$\begin{array}{|l|c|r|} \hline x & f(x)=\frac12 x^2 & (x,f(x))\\ \hline 0 & \frac12 (0)^2 & 0\\ \hline 2 & \frac12 (2)^2 & 2\\ \hline 4 & \frac12 (4)^2 & 8\\ \hline -2 & \frac12 (-2)^2 & 2\\ \hline -4 & \frac12 (-4)^2 & 8\\ \hline \end{array}$$

Let's look at these points on a graph.

We can already see the parabolic shape. Moving from left to right, the graph goes downward until $(0,0)$, then goes upwards. Let's connect these points.

There's our parabola!

He brings up a question in the end that is never answered.

What can you say about the graph of the function $f(x)=ax^2$ if $a$ is a negative number?

A negative $a$ value flips the graph vertically. We have the graph of $f(x)=\frac12 x^2$ above, so let's look at the graph of $f(x)= -\frac12 x^2$.

We see the points $(0,0), (-2,-2), (2,-2), (-4,-8),$ and $(4,-8)$. This means the parabola opens downward rather than upward.