write the equation in standard form and then graph the equation.
y=2x^2-8x+11
You will have to use "completing the square".
y = 2x^2-8x+11
= 2(x^2 - 4x+ .... ) + 11 <---- took out the coefficient of the square term from the first two terms
= 2(x^2 - 4x + 4 - 4) <--- took half of the coefficient of the x term, then squared it, added it and immediately subtracted it again, the net result was adding zero.
= 2( (x-2)^2 - 4) + 11 <----- formed the perfect square plus the left over constant
= 2(x-2)^2 - 8 + 11 <---- dististributed the 2 over the perfect square and the left-over constant
= 2(x-2)^2 + 3 <---- added the stuff at the end
all done, now I can read the vertex
which is .......
and if I let x = 0 in the original (or this last) equation, I get
y = 11
So plot the vertex, plot the y-intercept and form a neat parabola from the vertex to hit (0,11). Complete the other half of the parabola.
It should look like this:
http://www.wolframalpha.com/input/?i=plot+y+%3D+2x%5E2-8x%2B11,+y+%3D+2(x-2)%5E2+%2B+3
notice I graphed both forms of the equation and the graphs coincide. This was a neat way to check my work.
To write the equation y = 2x^2 - 8x + 11 in standard form, we need to rearrange the terms to have the highest power of x first, followed by the middle term, and then the constant term.
So, let's rearrange the equation:
y = 2x^2 - 8x + 11
Bring all terms to one side:
2x^2 - 8x + 11 - y = 0
Now, let's graph the equation.
To graph a quadratic equation like this, we need to find the vertex, y-intercept, and any x-intercepts.
The vertex formula for a quadratic equation in the form y = ax^2 + bx + c is given by:
x = -b / (2a)
y = c - b^2 / (4a)
So, in this case, a = 2, b = -8, and c = 11.
Calculating the x-coordinate of the vertex:
x = -(-8) / (2 * 2) = 8 / 4 = 2
Plugging the x-coordinate into the equation to find the y-coordinate of the vertex:
y = 2(2)^2 - 8(2) + 11
y = 8 - 16 + 11
y = 3
So, the vertex is (2, 3).
Now, let's find the y-intercept by plugging in x = 0 into the equation:
y = 2(0)^2 - 8(0) + 11
y = 0 - 0 + 11
y = 11
So, the y-intercept is (0, 11).
To find the x-intercepts, we need to solve the equation y = 2x^2 - 8x + 11 for x when y = 0:
0 = 2x^2 - 8x + 11
However, this equation does not factor easily, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -8, and c = 11.
Plugging these values into the quadratic formula:
x = (-(-8) ± √((-8)^2 - 4 * 2 * 11)) / (2 * 2)
x = (8 ± √(64 - 88)) / 4
x = (8 ± √(-24)) / 4
x = (8 ± 2√6i) / 4
Since the discriminant is negative, there are no real x-intercepts. The graph will not intersect the x-axis.
Finally, let's plot the points and graph the equation:
The vertex is (2, 3).
The y-intercept is (0, 11).
Since there are no x-intercepts, we won't be able to draw the parabola crossing the x-axis.
Therefore, the graph will be a downward-opening parabola passing through the vertex (2, 3) and y-intercept (0, 11).
To write the equation in standard form, we need to rearrange the equation so that the terms are in descending order of degrees with all the coefficients and constants on one side of the equation.
Given equation: y = 2x^2 - 8x + 11
To start, let's expand the equation:
y = 2x^2 - 8x + 11
Now, let's bring all the terms to one side of the equation:
2x^2 - 8x + (11 - y) = 0
Finally, let's rearrange the terms in descending order of degrees:
2x^2 - 8x + (11 - y) = 0
This equation is now in standard form.
To graph the equation, let's create a coordinate plane and plot some points.
First, let's find the vertex of the parabola using the formula: x = -b / (2a)
For our equation, a = 2 and b = -8, so
x = -(-8) / (2 * 2)
x = 8 / 4
x = 2
The x-coordinate of the vertex is 2. To find the y-coordinate, substitute x = 2 into the equation:
y = 2(2)^2 - 8(2) + 11
y = 2(4) - 16 + 11
y = 8 - 16 + 11
y = 3
So, the vertex is (2, 3).
Next, let's find a few more points to plot the graph. We can select some x-values and calculate the corresponding y-values using the equation.
When x = 0:
y = 2(0)^2 - 8(0) + 11
y = 0 - 0 + 11
y = 11
When x = 1:
y = 2(1)^2 - 8(1) + 11
y = 2 - 8 + 11
y = 5
When x = 3:
y = 2(3)^2 - 8(3) + 11
y = 18 - 24 + 11
y = 5
Finally, plot these points (0, 11), (1, 5), (2, 3), and (3, 5) on the graph and connect them with a smooth curve.
The graph of the equation y = 2x^2 - 8x + 11 will be a upward-opening parabola, passing through the points mentioned above. The vertex of the parabola is (2, 3).