Question

For each function, provide:

The concavity of the parabola.
The y-axis intercept.
The vertex coordinates.
The zeros of the function (if any).
a) f(x) = -x² - 3x + 4
b) g(x) = x² - 4
c) s(t) = 3t² - 6t + 3
d) p(n) = n² + 2n + 5
e) s(v) = -v² + 4v - 4
f) n(t) = -t² - 5

Answer 1

Concavity, y-axis intercepts, vertex coordinates, and zeros are determined for six given quadratic functions. The concavity direction depends on the leading coefficient, whereas vertex and intercepts are calculated using specific formulas and factoring or using the quadratic formula.

Step-by-step explanation:

We are examining various quadratic functions to determine their concavity, y-axis intercept, vertex coordinates, and zeros.

• a) f(x) = -x² - 3x + 4: The concavity is downwards since the coefficient of x² is negative. The y-axis intercept is at (0, 4), where x is 0. Using the vertex formula x = -b/2a, the vertex is at (-3/(-2), f(-3/(-2))). The zeros can be found by factoring or using the quadratic formula.
• b) g(x) = x² - 4: Concavity is upwards, y-intercept at (0, -4), vertex at (0, -4) as it's already in vertex form, and zeros at x = ±2 by factoring to (x-2)(x+2).
• c) s(t) = 3t² - 6t + 3: Upwards concavity, y-intercept at (0, 3), vertex found by the formula t = -(-6)/(2 * 3), and zeros found by the quadratic formula as it does not factor neatly.
• d) p(n) = n² + 2n + 5: Upwards concavity, y-intercept at (0, 5), vertex at n = -2/2(1), but no real zeros as the discriminant b² - 4ac is negative.
• e) s(v) = -v² + 4v - 4: Downwards concavity, y-intercept at (0, -4), vertex at v = -4/(-2), and zeros found by factoring or the quadratic formula.
• f) n(t) = -t² - 5: Downwards concavity, y-intercept at (0, -5), vertex at (0, -5), and no real zeros as the parabola does not cross the x-axis.