Question

what is the vertex of the quadratic function

\\(f(x) = 5(x - 4)(x + 2)\\)?

\\(\bigcirc\\ (-1, -25)\\)
\\(\bigcirc\\ (-4, 2)\\)
\\(\bigcirc\\ (1, -45)\\)
\\(\bigcirc\\ (4, -2)\\)

Explanation:

⚡ Using what you learned: vertex and axis of symmetry

Step 1: Find the x-intercepts

The quadratic function is given in factored form:

$$ f(x) = 5(x - 4)(x + 2) $$

Set each factor to zero to find the x-intercepts:

$$ x - 4 = 0 \implies x = 4 $$

$$ x + 2 = 0 \implies x = -2 $$

Step 2: Find the x-coordinate of the vertex

The axis of symmetry (and the x-coordinate of the vertex, \( h \)) lies exactly halfway between the two x-intercepts:

$$ h = \frac{4 + (-2)}{2} = \frac{2}{2} = 1 $$

Step 3: Find the y-coordinate of the vertex

Substitute \( x = 1 \) back into the original function to find the y-coordinate (\( k \)):

$$ f(1) = 5(1 - 4)(1 + 2) $$

$$ f(1) = 5(-3)(3) $$

$$ f(1) = -45 $$

The vertex is \( (1, -45) \).

Answer:

\( (1, -45) \)