Answer:
[tex]2(x-3)^2-15[/tex]
[tex]M =(-1, -15)[/tex]
Step-by-step explanation:
To express 2x² - 12x + 3 in the form a(x + b)² + c, where a, b and c are integers, we need to complete the square.
To complete the square, begin by factoring out the leading coefficient 2 from the x² term and the x term:
[tex]2(x^2 - 6x) + 3[/tex]
Now, add the square of half the coefficient of the x term inside the parentheses, and subtract the distributed value of this outside the parentheses:
[tex]2\left(x^2 - 6x+\left(\dfrac{-6}{2}\right)^2\right) + 3-2\left(\dfrac{-6}{2}\right)^2[/tex]
Simplify:
[tex]2\left(x^2 - 6x+\left(-3\right)^2\right) + 3-2\left(-3\right)^2\\\\\\2\left(x^2 - 6x+9\right) + 3-2(9)\\\\\\2(x^2-6x+9)+3-18\\\\\\2\left(x^2 - 6x+9\right) -15[/tex]
We have now created a perfect square trinomial inside the parentheses.
Factor the perfect square trinomial to complete the square:
[tex]2(x-3)^2-15[/tex]
Therefore, 2x² - 12x + 3 expressed in the form a(x + b)² + c is:
[tex]\Large\boxed{\boxed{2(x-3)^2-15}}[/tex]
[tex]\dotfill[/tex]
The quadratic curve C has the equation:
[tex]y = 2(x + 4)^2 - 12(x + 4) + 3[/tex]
As the leading coefficient is positive, the parabola opens upward, and it has a minimum point at its vertex.
If we compare the equation of curve C to the expression from the previous question, we can see that they are the same except that 4 has been added to the x variable. Since we can rewrite 2x² - 12x + 3 as 2(x - 3)² - 15, then we can rewrite y = 2(x + 4)² - 12(x + 4) + 3 as:
[tex]y = 2(x + 4 - 3)^2 - 15[/tex]
This simplifies to:
[tex]y = 2(x + 1)^2 - 15[/tex]
The vertex form of a quadratic equation is y = a(x - h)² + k, where (h, k) is the vertex of the parabola. So, in this case, the vertex is (-1, -15), which is the minimum point on C. Given that point M is the minimum point, then the coordinates of point M are:
[tex]\Large\boxed{\boxed{(-1,-15)}}[/tex]
