If each song was 2.5 minutes long, how many minutes of music total did Jen listen to?
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It is given that two vertices of square are (0,0) and (4,2).
Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.
1. By Supposing that these two are adjacent vertices of Square
The third vertex will be at (-4,2) which lies in third quadrant.
Suppose the coordinate of fourth vertex be (x,y).
Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).
Mid point of line joining (x,y) and (0,0) is (x/2,y/2).
Since diagonals of square bisect each other,
∵ x/2=0
⇒x=0
and
y/2=2
⇒y=4
So, The Coordinate of fourth vertex is (0,4).
Now coming back to second condition if these are two opposite vertex of Square.
Let the third coordinate be (a,b).
Length of diagonal=
Now,let side of Square be A.
Then length of Diagonal of square =√2 A
⇒√2 A=2√5
⇒A =√10
As third vertex is (a,b).
Using distance formula
a² + b²=10 -------------(1)
(a-4)²+ (b-2)²=10 --------------(2)
Solving expression (1) and (2), we get
⇒a²+ b²=(a-4)² +(b-2)²
⇒2a + b =5
⇒b=5-2a
Putting the value of b in (1),we get
⇒a² +(5-2a)²=10
⇒a²+25+4a²-20a =10
⇒5a²-20a+15=0
⇒a² - 4a + 3=0
Splitting the middle term,we get
⇒(a-3)(a-1)=0
⇒a=3 ∧ a=1
we get b=5-2×1=3 and b=5-2×3=5-6=-1
So,the other vertex are (1,3) and(3,-1).
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.
