Q: How could the function y=3t^2 +4 be plotted on a cartesian graph to produce a straight line? What would be the numerical values of the slope and the intercept of the line?

For this it's explained in the lab book for physics as if the function is y=mx+b it would be a straight line but if it's y=mx^2+b then it is a parabola. It's explained that to get a straight line I should go and make the equation with the square y=mx'+b with x'= x^2. Then it says that if you plotted these numbers with the corresponding y values you'd get a straight line.

I'm not sure I understand how this would work.

For the 2nd part, wouldn't the slope be the same as the original equation (3)? and the y intercept be the same as the original equation too (4)?

Thanks =D

Let x = t^2 and use equal increments of t^2 along the x axis, when plotting y vs x.

For t = 0, x = 0.
For t = 1, x = 1.
For t = 1.414, x = 2
For t = 1.732, x = 3 .
For t= 2, x = 4, etc.

The slope of the line you get will be 3.

Yes, when you plot y vs x (=t^2) . I forgot to mention the intercept. Sorry

Thanks DrWls =D

I have no idea how to do this

Ah, the wonders of math and graphs! Let me break it down for you, my friend.

To plot the function y = 3t^2 + 4 on a Cartesian graph to produce a straight line, you can indeed use a clever substitution. The idea is to introduce a new variable, let's call it x', which is equal to t^2. By doing so, we transform the equation into y = 3x' + 4, which does give you a straight line. Clever, right?

Now, to the numerical values of the slope and intercept of this line. In this transformed equation, y = 3x' + 4, the slope is indeed the coefficient in front of x', which is 3. So the slope is 3. Good job on catching that!

As for the y-intercept, it remains the same as in the original equation. So the y-intercept is still 4. Looks like you're on the right track!

So, to summarize:
- The slope of the line is 3 (same as the original equation, yay!)
- The y-intercept of the line is 4 (also the same as the original equation, double yay!)

Hope that clears things up for you! If you have any more math questions or need a good joke, feel free to ask. I'm here to help and make you smile!

To plot the function y = 3t^2 + 4 on a cartesian graph as a straight line, you can make a change of variables to transform it into the form y = mx + b, where m represents the slope and b represents the y-intercept.

Let's make the substitution x' = t^2. Now, let's rewrite the equation with the new variable:

y = 3t^2 + 4
y = 3(x') + 4

Now, if we plot the values of x' on the x-axis and the corresponding y-values on the y-axis, we will obtain a straight line. This is because x' is linear, which means the equation y = 3x' + 4 will produce a linear relationship between x' and y.

Regarding the numerical values of the slope and intercept, in the original equation y = 3t^2 + 4, the coefficient multiplied by t^2 represents the curvature of the parabola (not the slope). Therefore, in the transformed equation y = 3(x') + 4, the slope is given by the coefficient of x', which is 3. This means that for every unit increase in x', the corresponding y-value increases by 3 units.

As for the y-intercept, which is the value of y when x' (or t^2) is equal to zero, it remains the same as the original equation, which is 4. This is because the change of variables does not affect the y-intercept.

To summarize, the numerical values of the slope and y-intercept for the transformed equation y = 3(x') + 4 are a slope of 3 and a y-intercept of 4.

Okay thanks but wouldn't the y intercept still equal 4 DrWls?