How do you:
Complete the square as an aid in graphing.
[I put exponents in ().]
y = x(2) + 4x - 2
Please Help?
The complete square of the x^2 binomial is
x^2 + 4x + 4 = (x+2)^2
Therefore, add 6 to each side of the equation and you will have this still-valid equation with a perfect squate on the right side:
y + 6 = (x + 2)^2
The function will plot as a parabola rising above the y=6 line, tangent to it at x = -2.
To complete the square in order to aid in graphing, you can follow these steps:
1. Start with the given quadratic equation in the form y = ax^2 + bx + c. In this case, the equation is y = x^2 + 4x - 2.
2. Take the coefficient of the x term (in this case, 4) and divide it by 2, then square the result. This will give you (4/2)^2 = 2^2 = 4.
3. Add this result to both sides of the equation. So, add 4 to both sides of y = x^2 + 4x - 2.
y + 4 = x^2 + 4x + 4
4. The right side of the equation can now be factored as a perfect square. In this case, x^2 + 4x + 4 can be written as (x + 2)^2.
y + 4 = (x + 2)^2
5. If you want to isolate y to see the equation in vertex form, you can subtract 4 from both sides of the equation:
y = (x + 2)^2 - 4
Now, by completing the square, the equation y = x^2 + 4x - 2 has been transformed into y = (x + 2)^2 - 4, which is in vertex form. The vertex form allows you to easily identify the vertex of the parabola. In this case, the vertex is (-2, -4).
To graph the equation, plot the vertex (-2, -4) and draw a parabolic curve that opens upward or downward depending on whether the coefficient of x^2 is positive or negative. In this case, since the coefficient is positive, the parabola will open upward. The curve will be tangent to the line y = -4 at x = -2.