Given:
The vertex of a quadratic function is (4,-7).
To find:
The equation of the quadratic function.
Solution:
The vertex form of a quadratic function is:
...(i)
Where a is a constant and (h,k) is vertex.
The vertex is at point (4,-7).
Putting h=4 and k=-7 in (i), we get
The required equation of the quadratic function is where, a is a constant.
Putting a=1, we get
Therefore, the required quadratic function is .
Answer:
The solution of system of equation is (-2,0)
Step-by-step explanation:
Given system of equation are
Equation 1 : 2x+y=(-4)
Equation 2 : y+x=(-1)
To plot the equation of line, we need at least two points
For Equation 1 : 2x+y=(-4)
Let x=0
2x+y=(-4)
2(0)+y=(-4)
y=(-4)
Let x=1
2x+y=(-4)
2(1)+y=(-4)
y=(-6)
Therefore,
The required points for equation is (0,-4) and (1,-6)
For Equation 2 : y+x=(-1)
Let x=0
y+x=(-1)
y+(0)=(-1)
y=(-1)
Let x=2
y+x=(-1)
y+(2)=(-1)
y=(-2)
The required points for equation is (0,-1) and (2,-2)
Now, plot the graph using this points
From the graph,
The red line is equation 1 and blue line is equation 2
Since. The point of intersection is solution of system of equations
The solution of system of equation is (-2,0)
Answer:
Step-by-step explanation:
We want to calculate the right-endpoint approximation (the right Riemann sum) for the function:
On the interval [-1, 1] using five equal rectangles.
Find the width of each rectangle:
List the <em>x-</em>coordinates starting with -1 and ending with 1 with increments of 2/5:
-1, -3/5, -1/5, 1/5, 3/5, 1.
Since we are find the right-hand approximation, we use the five coordinates on the right.
Evaluate the function for each value. This is shown in the table below.
Each area of each rectangle is its area (the <em>y-</em>value) times its width, which is a constant 2/5. Hence, the approximation for the area under the curve of the function <em>f(x)</em> over the interval [-1, 1] using five equal rectangles is:
