PLEASE HELP ME!!! WILL MARK AS BRAINLIEST!!!!!!!
1 answer:
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The equation I picked was x^2+5x-6=3
1. the parabola opens up as the leading coefficient is a positive.
2. to find the vertex to find the x value
-5/2(1)= -5/2 (-2 1/2) (-2.5)
to find the y value of the vertex substitie the x value into the equation
(-2 1/2)^2+5(-2 1/2)-6=3
6 1/4 - 12 1/2 -9 = y
-15 1/4 = y
y = -15 1/4 (-15.25)
(-2.5, 15.25) , (-2 1/2, -15 1/4)
3. the y intercept can be found by substituting the X's with 0's
x^2+5x-6=3
first move the terms to the left
x^2+5x-9=y
(0)^2+5(0)-9=y
0+9=y
9=y
y=9
the y intercept is 9 (0,9)
4. b^2-4ac
5^2-4(1)(-9)
=25+36
=61
61 is greater than 0 so this means there are 2 real solutions.
5.
the +- next to each other is meant to be plus or minus
6. to find three other points pick any x value and substitute that into the equation
y=(2)^2+5(2)-9
y=4+10-9
y=5
(2,5)
y=(3)^2+5(3)-9
y=9+15-9
y=15
(3,15)
y=(4)^2+5(4)-9
y=16+20-9
y=27
(4,27)
I hope this helps, brainliest?
1. the parabola opens up as the leading coefficient is a positive.
2. to find the vertex to find the x value
-5/2(1)= -5/2 (-2 1/2) (-2.5)
to find the y value of the vertex substitie the x value into the equation
(-2 1/2)^2+5(-2 1/2)-6=3
6 1/4 - 12 1/2 -9 = y
-15 1/4 = y
y = -15 1/4 (-15.25)
(-2.5, 15.25) , (-2 1/2, -15 1/4)
3. the y intercept can be found by substituting the X's with 0's
x^2+5x-6=3
first move the terms to the left
x^2+5x-9=y
(0)^2+5(0)-9=y
0+9=y
9=y
y=9
the y intercept is 9 (0,9)
4. b^2-4ac
5^2-4(1)(-9)
=25+36
=61
61 is greater than 0 so this means there are 2 real solutions.
5.
the +- next to each other is meant to be plus or minus
6. to find three other points pick any x value and substitute that into the equation
y=(2)^2+5(2)-9
y=4+10-9
y=5
(2,5)
y=(3)^2+5(3)-9
y=9+15-9
y=15
(3,15)
y=(4)^2+5(4)-9
y=16+20-9
y=27
(4,27)
I hope this helps, brainliest?