To find the coordinates of the point on the parabola \(y = x^2 + 7x + 2\) that is nearest to the straight line \(y = 3x - 3\), we follow these steps:
- The distance \(D\) between a point \((x_1, y_1)\) and a line \(Ax + By + C = 0\) is given by the formula: \(D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\) Here, for the line \(y = 3x - 3\), we can rewrite it as \(3x - y - 3 = 0\). Thus, \(A = 3\), \(B = -1\), and \(C = -3\).
- The point on the parabola can be represented as \((x, x^2 + 7x + 2)\). Let's find the distance between this point and the line: \(D = \frac{|3x - (x^2 + 7x + 2) - 3|}{\sqrt{3^2 + (-1)^2}}\) Simplifying inside the absolute value: \(= \frac{|3x - x^2 - 7x - 2 - 3|}{\sqrt{10}}\) \(= \frac{| -x^2 - 4x - 5 |}{\sqrt{10}}\)
- To minimize the distance \(D\), we need to make \(-x^2 - 4x - 5\) as close to zero as possible, i.e., find the vertex of this quadratic expression. The expression \(-x^2 - 4x - 5\) can be written as \(-1 \cdot (x^2 + 4x + 5)\).
- The vertex of a quadratic \(ax^2 + bx + c\) is at \(x = -\frac{b}{2a}\). So for \(x^2 + 4x + 5\), the vertex is at: \(x = -\frac{4}{2 \cdot 1} = -2\)
- Substitute \(x = -2\) back into the parabola's equation to find \(y\): \(y = (-2)^2 + 7(-2) + 2 = 4 - 14 + 2 = -8\)
- Thus, the coordinates of the point on the parabola nearest to the line are \((-2, -8)\).
Therefore, the correct answer is \((-2, -8)\).