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how do i do these, may someone show me how please :)
1.Give the turning point of the following quadratic:
y = ( 6+5 x ) ( 4 x+1 )
2.Give the turning point of the following quadratic
y = -9 x2+2 x+1
are you in Calculus? or pre-calculus?
methods are VERY different.
Basically, the turning point is the vertex of the parabola.
Find the vertex:
on (1) it is half way between the zeros, so average the two zeros. (-6/5 + -1/4) / 2 = -29 / 40
x = -29/40
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on(2) use vertex ==> x=-b / 2a = -2 / (2(-9)) = 1/9
x = 1/9
If you want the point, you must plug each x into the corresponding y= to get the (x,y) point
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If you are in calculus, set the first derivaitve = 0 to find the x where it turns around
40x + 29 = 0
and
-18x+2 = 0
y = (6 + 5x)(4x + 1) = (5x + 6)(4x + 1)
This is a parabola that opens upward. The only turning point is the vertex:
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The x value of the vertex is midway between the x-intercepts:
5x + 6 = 0
5x = -6
x = -6/5
4x + 1 = 0
4x = -1
x = -1/4
midpoint is the average of the 2 x values:
(-6/5 + -1/4 )/2 = ((-24 - 5)/20)/2 = (-29/20)/2 = -29/40
x = -29/40
y = (5(-29/40) + 6)(4(-29/40) + 1)
= (-145/40 + 6)(-58/10 + 1)
= (-145/40 + 240/40)(-58/10 + 10/10)
= (95/40)(-48/10)
= (19/8)(-24/5)
= (19)(-3/5)
= -57/5
Turning point = vertex = (-29/40, -57/5)
y = -9x^2 + 2x + 1
Parabola opening downward.
Vertex is the only turning point.
y = -(9x^2 - 2x - 1)
Find x-intercepts:
-(9x^2 - 2x - 1) = 0
9x^2 - 2x - 1 = 0
a = 9, b = -2, c = -1
x = (2 +/- sqrt((-2)^2 - 4(9)(-1)))/2(9)
x = (2 +/- sqrt(4 + 36))/18
x = (2 +/- sqrt(40))/18
x = (2 +/- 2sqrt(10))/18
x = (1 +/- sqrt(10))/9
Average of x-intercept gives x-coordinate of vertex:
[(1/9 + sqrt(10)/9) + (1/9 - sqrt(10)/9)]/2
= (2/9)/2
= 1/9
y = -(9x^2 - 2x - 1)
y = -(9(1/9)^2 - 2(1/9) - 1)
y = -(1/9 - 2/9 - 1)
y = -(-10/9)
y = 10/9
Vertex = turning point = (1/9, 10/9)
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