Parabola in Learning Program Shenanigans

[Iceycold12]

When to use the various Quadratic Formulas (Roots, Parabola, Actual Formula)

Hello everyone.

So I understand how to graph parabola's and what not, but this internet program my teacher provides makes everything more complicated 10 fold. So here's the problem:

http://gyazo.com/c441223b55d659349c75b9d8d728377f (screenshot)

Basically on the left side I need to plug in those numbers 1, then 3, into the x's in -16x^2+60x+6. But then they flip it around in Question 3 & 4, I need to figure it out when height is given, not time. The little tooltip says to set them equal but I don't understand that portion.

Here's an example of it done correctly: http://gyazo.com/df7cf1785ed3ea1dca4fffdafdacc4d9

See for question 3 & 4 it was done? Not sure on that. It doesn't' explain it well enough for me to understand what I need to do.

Here's a screenshot of the example done right but the WHOLE thing. http://gyazo.com/97483a86867c08ab26f7ea9f82e32606

Sorry for making two threads already so soon but I can't get past this part of the program, like I stated before I know how to graph parabola's but this gives real life examples and it makes it a bit more confusing especially on those parts I showed.

 

[wjm11]

Basically on the left side I need to plug in those numbers 1, then 3, into the x's in -16x^2+60x+6. But then they flip it around in Question 3 & 4, I need to figure it out when height is given, not time. The little tooltip says to set them equal but I don't understand that portion.

What they mean is to set up the equation: y = -16x^2+60x+6, where y is height in feet. So, for your problem, you would substitute the height, 56 ft, in place of y and solve for x (time in seconds):

56 = -16x^2+60x+6

Have you learned how to solve equations like this (quadratics)? Are you familiar with the quadratic formula?

Graphically, you would first draw the parabola; that represents the entire y = -16x^2+60x+6 for any x or y. Next, if you are interested in 56 ft, draw a horizontal line at y = 56 through the parabola. You'll see it crosses the parabola at two spots. The x values of those spots are the times the object is at 56 feet. Make sense?

 

[Iceycold12]

I know how to solve ax^2+bx+c equations well not sure if it means solving but I can find the vertex axis of symettry basically how to find the (x,y) of that ax^2+bx+c and graph the parabola. (-b/2a & replacing x for 0 to find y intercept all those messy calculations)

But that you showed me 56 = -16x^2+60x+6 I have not learned yet, in fact I just started what I stated above this morning. But the teacher assigns this unit without reading the content inside.

I googled "solving quadratic equations" but they have it set up as two binomials and they don't have 56 = ax^2+bx+c format. What do I need to do here to solve it?

 

[Iceycold12]

Got it thanks so much.

The difference between using the parent formula for graphing y=x^2 (where I would make an x,y table to determine points) and the actual quadratic formula ax^2+bx+c = 0 is that the quadratic formula requires ax^2 bx and c to execute right?

To better explain this, I got a worksheet asking to graph the function f(x)=5x^2 or y=5x^2. So I made a small table. (input -> output)

0->0 1->5 2->20 3-> 45 -1 -> 5 -2 -> 20 -3 -> 45. Then I set my vertex at 0,0 and plotted the points (1,5) (2,20) and so on. What I mean is that I didn't use ax^2+bx+c because I never had a bx or c term, so I'm assuming this is correct?

 

[Iceycold12]

Hey guys, on top of what I said above, I learned how to solve a quadratic equation using roots or the actual equation.

There's so many quadratic stuff to remember, that I know well, but not sure when to use which method.

Like, when do I know I will use the method in which I find the vertex, axis of symmetry and y-intercept when given something like y = -2x-8x-3 (The AX^2 + BX + C situation.

And when do I know I need to solve with a quadratic formula? -b + or - square root b^2-4ac/2a (sorry don't know how to type it well on keyboard)

Oh and 1 more thing do ALL the quadratic equations end in =0 at the end, I don't see it sometimes, should I always assume it's there?

Thanks.

 

[srmichael]

Hey guys, on top of what I said above, I learned how to solve a quadratic equation using roots or the actual equation.

There's so many quadratic stuff to remember, that I know well, but not sure when to use which method.

Like, when do I know I will use the method in which I find the vertex, axis of symmetry and y-intercept when given something like y = -2x-8x-3 (The AX^2 + BX + C situation.

And when do I know I need to solve with a quadratic formula? -b + or - square root b^2-4ac/2a (sorry don't know how to type it well on keyboard)

Oh and 1 more thing do ALL the quadratic equations end in =0 at the end, I don't see it sometimes, should I always assume it's there?

Thanks.

When you "solve" a quadratic function, you are finding the x-intercepts (also known as roots or zeros) of the equation. To find the x-intercepts, you set y = 0, that is why you always see the quadratic function = 0 when you "solve" it.

 

[Iceycold12]

So then what's this for?

Find the vertex, axis of symmetry and y-intercept when given something like y = -2x-8x-3 (The AX^2 + BX + C) situation.

I'm guessing the above is when they ask u to draw one and "solve" means use the formula? What if you solve a formula and they tell u to graph it, put the two x points and find the vertex by -b/2a (find axis of symmetry afterwards)? I'm guessing questions would be worded differently on an exam for this?

edit: Bah, here's a picture of what I mean by "Find the vertex, axis of symmetry and y-intercept when given something like y = -2x-8x-3 (The AX^2 + BX + C) situation" it's my classwork I want to know when I do this, and when I need to just solve: (Sorry for breaking your neck, picture was horizontal for some odd reason).

edit2: Do the quadratic formula and this screenshot below the same? Like do they result in the same answer?

 

[Iceycold12]

Okay so today I learned the discriminant formula today which is the

, and all this quadratic stuff is getting tangled.

b^2-4ac > 0
b^2-4ac < 0
b^2-4ac = 0

Which means: If I apply that formula to say

, I would get -8 which means 0 roots. The full rules are:
Positive Number When Solved = 2 roots
Negative Number When Solved = 0 roots
Zero When Solved = 1 root

Now, do I got this right (look below)

Use the discriminant when asked for how many roots a given quadratic equation has
Use the quadratic formula to solve the two roots of a quadratic equation
When ASKED to find the vertex, axis of symmetry, and the y-intercept of a parabola, use the -b/2a, plugin x find the (x,y) of the vertex and then use the c value as the intercept, do draw the parabola

OKAY, so assuming all this is right, what I said above, does this mean the last italics line is pointless after I know the quadratic formula? Can't I just find the two x intercept cross points (roots) and then apply -b/2a to find the axis of symmetry and draw the parabola?

Sigh, I have a big weight on my shoulders with this.

 

[wjm11]

Okay so today I learned the discriminant formula today which is the b^2-4ac, and all this quadratic stuff is getting tangled.

b^2-4ac > 0
b^2-4ac < 0
b^2-4ac = 0

Which means: If I apply that formula to say x^2-6x+11, I would get -8 which means 0 roots. The full rules are:
Positive Number When Solved = 2 roots
Negative Number When Solved = 0 roots
Zero When Solved = 1 root

Now, do I got this right (look below)

Use the discriminant when asked for how many roots a given quadratic equation has
Use the quadratic formula to solve the two roots of a quadratic equation
When ASKED to find the vertex, axis of symmetry, and the y-intercept of a parabola, use the -b/2a, plugin x find the (x,y) of the vertex and then use the c value as the intercept, do draw the parabola

OKAY, so assuming all this is right, what I said above, does this mean the last italics line is pointless after I know the quadratic formula? Can't I just find the two x intercept cross points (roots) and then apply -b/2a to find the axis of symmetry and draw the parabola?

Actually, all these pieces of information fit together nicely.

The discriminant gives us some general info about how the graph will look in advance. If it’s +, the parabola crosses the x-axis at two places. If it’s 0, the vertex of the parabola lies on the x-axis. If it’s -, the parabola does not cross the x-axis.

The quadratic formula tells us where the parabola crosses (or touches) the x-axis.

The “-b/2a” approach let’s us find the vertex, and “c” tells us the y-intercept.

Make sense?

 

[Iceycold12]

Yeah I see what you mean there, thanks.

But what I'm trying to ask is in which scenarios do you use which method?

Can I just ignore the:

When ASKED to find the vertex, axis of symmetry, and the y-intercept of a parabola, use the -b/2a, plugin x find the (x,y) of the vertex and then use the c value as the intercept, do draw the parabola

And just incorporate the whole draw a parabola with the quadratic formula? Since it gives two x points, then the axis of symmetry shows the exact middle of those two points and then draw the curved parabola lines?

If you don't know what I mean by

When ASKED to find the vertex, axis of symmetry, and the y-intercept of a parabola, use the -b/2a, plugin x find the (x,y) of the vertex and then use the c value as the intercept, do draw the parabola

Scroll up to where my picture is. On the end of page 1.

I understand all the concepts. I'm just confused on when to use what, since there's so many formulas.

 

[wjm11]

Can I just ignore the: When ASKED to find the vertex, axis of symmetry, and the y-intercept of a parabola, use the -b/2a, plugin x find the (x,y) of the vertex and then use the c value as the intercept, do draw the parabola


And just incorporate the whole draw a parabola with the quadratic formula? Since it gives two x points, then the axis of symmetry shows the exact middle of those two points and then draw the curved parabola lines?

No. What if your parabola does not have any x-axis intercepts? The quadratic formula will not help you graph it then.

All parabolas have a vertex and a y-intercept. The "-b/2a" and "c" approach will allow you to find those points. From the vertex and y intercept, you can tell if the parabola will cross the x-axis or not. Then use the quadratic formula if appropriate.

 

[Iceycold12]

No. What if your parabola does not have any x-axis intercepts? The quadratic formula will not help you graph it then.

All parabolas have a vertex and a y-intercept. The "-b/2a" and "c" approach will allow you to find those points. From the vertex and y intercept, you can tell if the parabola will cross the x-axis or not. Then use the quadratic formula if appropriate.

Okay so if I understood right, it depends on what they ask on an exam?

So this orignally was right, the conclusion I arrived to:

Use the discriminant when asked for how many roots a given quadratic equation has
Use the quadratic formula to solve the two roots of a quadratic equation
When ASKED to find the vertex, axis of symmetry, and the y-intercept of a parabola, use the -b/2a, plugin x find the (x,y) of the vertex and then use the c value as the intercept, do draw the parabola

 

[wjm11]

So this orignally was right, the conclusion I arrived to:

Use the discriminant when asked for how many roots a given quadratic equation has
Use the quadratic formula to solve the two roots of a quadratic equation
When ASKED to find the vertex, axis of symmetry, and the y-intercept of a parabola, use the -b/2a, plugin x find the (x,y) of the vertex and then use the c value as the intercept, do draw the parabola

Looks good. Well done.

 

[Iceycold12]

Thank you so much, feel much more comfortable now.

Just curious, could you apply -b/2a (axis of symmetry) to the quadratic equation AX^2 + BX + C = 0 and draw a parabola, or can you only find roots with quadratic formula?