I am having a bit of trouble with this math. my teacher has taught us how to graph quadratic functions differently than the tutorials all of the internet did. it involved taking "C" away, then factoring, splitting "B" in half, and a bunch of stuff, sorry im not being too specific, my teacher told us not to take notes because we would get lost and now im stuck at home stressin' out over this homework. so does anyone know how to solve these problems "my way" a step by step would be very helpful. thanks for reading.
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I don't know what your way is.
But, let's see if this helps you:
ALL quadratic functions graph as parabolas.
The key points to plot are the vertex and the x-intercepts (if they exist).
➊ Quadratic functions have the form y=ax²+bx+c.
➋ The x-coordinate of the vertex is determined by x= -b/(2a).
The y-coordinate of the vertex is determined by plugging -b/(2a) in for x.
Another way to get the y-coordinate of the vertex is to
multiply the x-coordinate by b/2 then add c.
➌ If the coefficient of x² is positive, then the parabola opens upward.
If the coefficient of x² is negative, then the parabola opens downward.
➍ On the coordinate system, y = 0 is the x-axis.
So, the x-intercepts are found by setting y equal to 0 and then solving for x.
➎ Notice that when x=0 we have that y = c so that the y-intercept is c.
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That looks like a lot, but, it really isn't.
For example:
➊ y = -x²-4x+12 ← quadratic function ... a= -1, b= -4, c= +12 ... graphs as a parabola
➋ x-coordinate of the vertex = -b/(2a) = -(-4)/(-2) = -2
y-coordinate of vertex = (x-coordinate)(b/2) + c = -2(-4/2) + 12 = 16
OR } vertex V(-2,16)
y-coordinate of vertex = -x²-4x+12 = -(-2)²-4(-2)+12 = -4+8+12 = 16
➌ The coefficient of x² is -1 which is negative ... So, the parabola opens down.
➍ y = -x²-4x+12 ← set y to 0 and solve for x
0 = -x²-4x+12 ← get the coefficient of x² to +1 by factoring out -1
0 = -(x²+4x-12) ← Now, factor what's in parentheses
0 = -(x+6)(x-2) ← solutions are x= -6 and x=2
x-intercepts are at x = -6 and 2.
➎ Optional, the y-intercept = -x²-4x+12 = -0²-4·0+12 = 12
Hope that gives you enough info.
Have a good one!
——————————————————————————————————————
I don't know what your way is.
But, let's see if this helps you:
ALL quadratic functions graph as parabolas.
The key points to plot are the vertex and the x-intercepts (if they exist).
➊ Quadratic functions have the form y=ax²+bx+c.
➋ The x-coordinate of the vertex is determined by x= -b/(2a).
The y-coordinate of the vertex is determined by plugging -b/(2a) in for x.
Another way to get the y-coordinate of the vertex is to
multiply the x-coordinate by b/2 then add c.
➌ If the coefficient of x² is positive, then the parabola opens upward.
If the coefficient of x² is negative, then the parabola opens downward.
➍ On the coordinate system, y = 0 is the x-axis.
So, the x-intercepts are found by setting y equal to 0 and then solving for x.
➎ Notice that when x=0 we have that y = c so that the y-intercept is c.
——————————————————————————————————————
That looks like a lot, but, it really isn't.
For example:
➊ y = -x²-4x+12 ← quadratic function ... a= -1, b= -4, c= +12 ... graphs as a parabola
➋ x-coordinate of the vertex = -b/(2a) = -(-4)/(-2) = -2
y-coordinate of vertex = (x-coordinate)(b/2) + c = -2(-4/2) + 12 = 16
OR } vertex V(-2,16)
y-coordinate of vertex = -x²-4x+12 = -(-2)²-4(-2)+12 = -4+8+12 = 16
➌ The coefficient of x² is -1 which is negative ... So, the parabola opens down.
➍ y = -x²-4x+12 ← set y to 0 and solve for x
0 = -x²-4x+12 ← get the coefficient of x² to +1 by factoring out -1
0 = -(x²+4x-12) ← Now, factor what's in parentheses
0 = -(x+6)(x-2) ← solutions are x= -6 and x=2
x-intercepts are at x = -6 and 2.
➎ Optional, the y-intercept = -x²-4x+12 = -0²-4·0+12 = 12
Hope that gives you enough info.
Have a good one!
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