For a party, a student can spend no more than $15.00 on soda. Coca-cola cost $2.00 each and mountain dew cost $3.00 each.write t
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First lets see the pythagorean identities
So if we have to solve for sin theta , first we move cos theta to left side and then take square root to both sides, that is
Now we need to check the sign of sin theta
First we have to remember the sign of sin, cos , tan in the quadrants. In first quadrant , all are positive. In second quadrant, only sin and cosine are positive. In third quadrant , only tan and cot are positive and in the last quadrant , only cos and sec are positive.
So if theta is in second quadrant, then we have to positive sign but if theta is in third or fourth quadrant, then we have to use negative sign .
1. f(x)=x²+10x+16
Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(As, a>0 the parabola is open upward), b=10. by putting the values.
-b/2a = -10/2(1) = -5
f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9
So, Vertex = (-5, -9)
Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).
Now find x-intercept put y=0 in the above equation. 0= x²+10x+16
x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2
From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.
2. f(x)=−(x−3)(x+1)
By multiplying the factors, the general form is f(x)= -x²+2x+3.
Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(As, a<0 the parabola is open downward), b=2. by putting the values.
-b/2a = -2/2(-1) = 1
f(-b/2a)= f(1)=-(1)²+2(1)+3= 4
So, Vertex = (1, 4)
Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).
Now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.
-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1
From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.
3. f(x)= −x²+4
Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(As, a<0 the parabola is open downward), b=0. by putting the values.
-b/2a = -0/2(-1) = 0
f(-b/2a)= f(0)= −(0)²+4 =4
So, Vertex = (0, 4)
Now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).
Now find x-intercept put y=0 in the above equation. 0= −x²+4
−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2
From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.
4. f(x)=2x²+16x+30
Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(As, a>0 the parabola is open upward), b=16. by putting the values.
-b/2a = -16/2(2) = -4
f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2
So, Vertex = (-4, -2)
Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).
Now find x-intercept put y=0 in the above equation. 0=2x²+16x+30
2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3
From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.
5. y=(x+2)²+4
The general form of parabola is y=a(x-h)²+k , where vertex = (h,k)
if a>0 parabola is opened upward.
if a<0 parabola is opened downward.
Compare the given equation with general form of parabola.
-h=2 ⇒h=-2
k=4
so, vertex= (-2, 4)
As, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.
The graph is attached below.
