Give the coordinates of the vertex and graph the equation in a window that includes the vertex. y=(x+8)²-9.
I know how to graph just not how to do the problem. HELP!
You need to learn this quickly: this is NOT precalculus, but at best ordinary Alg II.
when x=-8, y is a minimum, so x=-8 must be at the vertex. When x=-8, y=-9, so (-8,-9) is the vertex.
Find the exact solutions to the quadratic equation in the complex numbers.
(x-13)²=-36
Kevin Smith I have answered that question three times for you already.
Posted by Kevin Smith on Friday, July 1, 2011 at 2:38pm.
Find the exact solutions to the quadratic equation in the complex numbers.
(x-13)²=-36
algebra - Damon, Friday, July 1, 2011 at 5:35pm
x - 13 = + sqrt (-36)
x = 13 + 6 i
or
x-13 = - sqrt (-36)
x = 13 - 6 i
Note that those two solutions are complex conjugates.
To find the vertex of the equation y = (x + 8)² - 9, we can use the formula x = -b / 2a, where a, b, and c are the coefficients of the quadratic function in standard form (y = ax² + bx + c).
In this case, the coefficient of x² is 1, the coefficient of x is 16 (since 16 * (x + 8)² = 16x² + 128x + 256), and the constant term is -9.
Using the formula, x = -b / 2a, we substitute the values: x = -16 / (2 * 1) = -16 / 2 = -8.
The x-coordinate of the vertex is -8.
To find the y-coordinate, substitute the x-coordinate back into the equation: y = (-8 + 8)² - 9 = 0 - 9 = -9.
So, the vertex of the quadratic equation is (-8, -9).
To graph the equation, you can use the vertex and a few additional points. Let's choose some x-values to determine the corresponding y-values:
When x = -10: y = (-10 + 8)² - 9 = (-2)² - 9 = 4 - 9 = -5. This gives us the point (-10, -5).
When x = -6: y = (-6 + 8)² - 9 = 2² - 9 = 4 - 9 = -5. This gives us the point (-6, -5).
Now that we have the vertex (-8, -9) and two additional points (-10, -5) and (-6, -5), we can plot them on a graph and draw a smooth curve passing through these points to represent the graph of the equation y = (x + 8)² - 9.