Format of Quadratic Equation: y = ax2 + bx + c
Given Quadratic Equation: y = 2x2 - 3x + 3
Coefficient Variable Values: a = 2 and b = -3 and c = 3
Axis of Symmetry: x = -b/2a = -(-3)/2(2) so answer is x = 3/4
Vertex: x value is axis of symmetry (3/4) and y value is calculated substituting 3/4 for x in original equation: y = 2(3/4)2 - 3(3/4) + 3 = 2(9/16) - 9/4 + 3 = 9/8 - 9/4 + 3 = 9/8 - 18/8 + 24/8 = 15/8,
so answer is (3/4,15/8)
x intercepts (solve using quadratic formula): x = (-b plus or minus sqrt(b2 - 4ac)/2a, so plugging in coefficient values for a and b and c, we get x = [-(-3) plus or minus sqrt((-3)2 - 4(2)(-3)]/2(2), which results in x = (3 + sqrt(33))/4 or (3 - sqrt(33))/4 and answers to nearest tenth are x = (3 + 5.7) / 4 = 2.2
or x = (3 - 5.7) / 4 = -0.7
y intercept is calculated by substituting zero for x into original equation: y = 2x2 - 3x + 3, so y-intercept is 3.
Domain is range is from calculated negative x intercept (-0.7) to calculated positive x intercept (2.2) and is written as (-0.7,2.2)
Range is from calculated y-intercept to positive infinity, since parabola opens up due to positive x2 coefficient value, so range is written as (3,positive infinity). Note: infinity symbol is sideways 8.
A cat is measured by millimeters, feet, inches
A.All squares of 4 right angles
B.All parallelograms have 4 right angles
<h3>
Answer: B) Only the first equation is an identity</h3>
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I'm using x in place of theta. For each equation, I'm only altering the left hand side.
Part 1
cos(270+x) = sin(x)
cos(270)cos(x) - sin(270)sin(x) = sin(x)
0*cos(x) - (-1)*sin(x) = sin(x)
0 + sin(x) = sin(x)
sin(x) = sin(x) ... equation is true
Identity is confirmed
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Part 2
sin(270+x) = -sin(x)
sin(270)cos(x) + cos(270)sin(x) = -sin(x)
-1*cos(x) + 0*sin(x) = -sin(x)
-cos(x) = -sin(x)
We don't have an identity. If the right hand side was -cos(x), instead of -sin(x), then we would have an identity.
