Master the formulas and techniques for finding the vertex, axis of symmetry, and direction of a parabola directly from its standard form equation, y = ax^2 + bx + c.
Given a parabola in standard form \(y = ax^2 + bx + c\), what is the formula for the x-coordinate of its vertex?
The x-coordinate of the vertex is given by the formula \(x = -\frac{b}{2a}\).
How do you find the y-coordinate of the vertex for a parabola in standard form, \(y = ax^2 + bx + c\), after finding the x-coordinate?
Substitute the x-coordinate of the vertex (\(x = -\frac{b}{2a}\)) back into the original equation \(y = ax^2 + bx + c\) and solve for \(y\).
Find the x-coordinate of the vertex for the parabola \(y = 2x^2 - 8x + 5\).
Using the formula \(x = -\frac{b}{2a}\) with \(a = 2\) and \(b = -8\), we get \(x = -\frac{-8}{2(2)} = \frac{8}{4} = 2\). The x-coordinate is \(2\).
Find the full vertex coordinates for the parabola \(y = 2x^2 - 8x + 5\).
The x-coordinate is \(x = 2\). Substitute this into the equation: \(y = 2(2)^2 - 8(2) + 5 = 2(4) - 16 + 5 = 8 - 16 + 5 = -3\). The vertex is \((2, -3)\).
What is the equation for the axis of symmetry of a parabola in standard form \(y = ax^2 + bx + c\)?
The axis of symmetry is a vertical line passing through the vertex. Its equation is the same as the formula for the x-coordinate of the vertex: \(x = -\frac{b}{2a}\).
What is the axis of symmetry for the parabola \(y = -x^2 + 6x - 1\)?
Using \(x = -\frac{b}{2a}\) with \(a = -1\) and \(b = 6\), we get \(x = -\frac{6}{2(-1)} = -\frac{6}{-2} = 3\). The axis of symmetry is \(x = 3\).
How do you determine if a parabola in standard form \(y = ax^2 + bx + c\) has a minimum or maximum value?
It depends on the sign of \(a\). If \(a > 0\), the parabola opens up and has a minimum value. If \(a < 0\), it opens down and has a maximum value. The value itself is the y-coordinate of the vertex.
Find the minimum value of the function \(f(x) = 3x^2 + 12x + 7\).
First, find the vertex. The x-coordinate is \(x = -\frac{12}{2(3)} = -2\). The y-coordinate (the minimum value) is \(f(-2) = 3(-2)^2 + 12(-2) + 7 = 12 - 24 + 7 = -5\). The minimum value is \(-5\).
What is the maximum value of the function \(g(x) = -2x^2 - 4x + 1\)?
The parabola opens down, so it has a maximum. The x-coordinate of the vertex is \(x = -\frac{-4}{2(-2)} = -1\). The maximum value is \(g(-1) = -2(-1)^2 - 4(-1) + 1 = -2 + 4 + 1 = 3\).
The sum of the roots (x-intercepts) of a quadratic \(ax^2+bx+c=0\) is \(-\frac{b}{a}\). How does this relate to the vertex?
The x-coordinate of the vertex, \(-\frac{b}{2a}\), is exactly half the sum of the roots. This means the vertex is always horizontally midway between the two x-intercepts of the parabola.
