Final answer:
To find the maximum/minimum value of the quadratic equation 2x²-8x-2, calculate the x-coordinate of the vertex by using the formula x = -b/2a, and then substitute the value of x back into the equation to find the y-coordinate. The maximum/minimum value is -10.
Step-by-step explanation:
To find the maximum/minimum value of the quadratic equation 2x²-8x-2, we can use the concept of vertex. The vertex of a quadratic equation in the form y = ax² + bx + c is given by the formula x = -b/2a. In this case, a = 2 and b = -8. Plugging in these values, we get x = -(-8)/(2*2) = 2.
The maximum/minimum value can be found by substituting the value of x back into the equation to find y. Evaluating 2(2)² - 8(2) - 2, we get y = 8 - 16 - 2 = -10.
To find the maximum or minimum value of the quadratic function 2x²-8x-2, we first need to find the vertex of the parabola represented by this function. The quadratic equation can be written in the form ax² + bx + c = 0, where a=2, b=-8, and c=-2.
The x-coordinate of the vertex can be found using the formula -b/(2a). Substituting our values, we get x = 8/(2*2) = 2. To find the y-coordinate of the vertex, we substitute x back into the original equation to get y = 2(2)2 - 8(2) - 2 = -10. Therefore, the vertex is at (2, -10), and since the coefficient a is positive, the parabola opens upwards, making the vertex a minimum point. Hence, the minimum value of the quadratic function is -10.
Therefore, the maximum/minimum value of the equation 2x²-8x-2 is -10.