Factoring the equation gives: (X + 1)^2 = 0 (or (X+ 1)(X + 1) = 0, so Y = 0 for X = - 1. That is of course the 'vertex'. Then you can plug in every number for X in the function f(X) = X^2 + 2X + 1. Then you also see the graphic.
equation is in proper quadratic form where y = ax² + bx + c. a =1, b = 2 and c =1. Because the a term is positive we have a "smiley face" parabola. The y intercept is always the c term which is [0 , 1] in this case (This is an x,y coordinate here.) Now for the vertex and the x intercepts. vertex form is defined as y = a(x - h)² + k where h is found by the formula h = 1/2(-b/a) and the k variable is found by the formula k = -a(h²) + c (NOTE the negative sign in front of the a term) h = 1/2(-2/1) which is -1 k = -1(h²) + 1 which is 0 y = 1(x + 1)² + 0 x = h + √(-k/a) and x = h - √(-k/a) [ ( ) are used to show both the k and a terms are under the square root sign] x = -1 + √(0/1) and x = -1 - √(0/1) or x =-1 and x = -1 Therefore the x intercept is the xy coordinate [-1,0] Super simple, and no need to use the quadratic formula for this type of problem. ( I haven't worked with higher than quadratic equations, so excuse me if stupid, but why the clunky and hard to remember quadratic formula is even used is beyond me. If the equation is in the "proper" form then the equations I used above are easier to remember and less clunky. edit: lol forgot the vertex it is also the xy coordinate of [-1,0] , but that is just a coincidence that was resulted from the coefficients used. Now you should have enough information to graph the equation.
Interesting, In German we call it a turning point (umkerpunkt translated). Never heard this as vertex in English, need to look into some language roots. No pun intended 😅
This is not Kindergarten! This is used in manufacturing. The parabola is the shape of lenses or mirrors of telescopes and microscopes and such things as the lens in the "LASER" in the optical drive of your computer. Could you also show how to find the focal point of a parabola? I once wrote a small BASIC program on my old Atari computer that figured the quadratic equation and plotted the graph on the screen for whatever x and y points I wanted. (it was saved on cassette tape, believe it or not)
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Factoring the equation gives: (X + 1)^2 = 0 (or (X+ 1)(X + 1) = 0, so Y = 0 for X = - 1. That is of course the 'vertex'. Then you can plug in every number for X in the function f(X) = X^2 + 2X + 1. Then you also see the graphic.
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Thank you!
Awesome
equation is in proper quadratic form where y = ax² + bx + c. a =1, b = 2 and c =1. Because the a term is positive we have a "smiley face" parabola. The y intercept is always the c term which is [0 , 1] in this case (This is an x,y coordinate here.)
Now for the vertex and the x intercepts.
vertex form is defined as y = a(x - h)² + k where h is found by the formula h = 1/2(-b/a) and the k variable is found by the formula k = -a(h²) + c (NOTE the negative sign in front of the a term)
h = 1/2(-2/1) which is -1
k = -1(h²) + 1 which is 0
y = 1(x + 1)² + 0
x = h + √(-k/a) and x = h - √(-k/a) [ ( ) are used to show both the k and a terms are under the square root sign]
x = -1 + √(0/1) and x = -1 - √(0/1) or x =-1 and x = -1
Therefore the x intercept is the xy coordinate [-1,0]
Super simple, and no need to use the quadratic formula for this type of problem. ( I haven't worked with higher than quadratic equations, so excuse me if stupid, but why the clunky and hard to remember quadratic formula is even used is beyond me. If the equation is in the "proper" form then the equations I used above are easier to remember and less clunky.
edit:
lol forgot the vertex it is also the xy coordinate of [-1,0] , but that is just a coincidence that was resulted from the coefficients used. Now you should have enough information to graph the equation.
Interesting, In German we call it a turning point (umkerpunkt translated). Never heard this as vertex in English, need to look into some language roots. No pun intended 😅
Nice 👍
This is not Kindergarten! This is used in manufacturing. The parabola is the shape of lenses or mirrors of telescopes and microscopes and such things as the lens in the "LASER" in the optical drive of your computer.
Could you also show how to find the focal point of a parabola?
I once wrote a small BASIC program on my old Atari computer that figured the quadratic equation and plotted the graph on the screen for whatever x and y points I wanted. (it was saved on cassette tape, believe it or not)
Let's solve a linear eqTion.
🙃
nice parab
I don't understand parabola
Is this kindergarden?