Question

find the smallest y value for the following quadratic equation. enter $-infty$ if there is no smallest value, i.e. the function decreases without bound.
$y = -2x^2 - 8x - 1$
show your work here
hint: to add infinity (\∞\), type \infinity\

Explanation:

Step1: Analyze the quadratic function

The quadratic function is given by \( y = -2x^2 - 8x - 1 \). For a quadratic function in the form \( y = ax^2 + bx + c \), the coefficient of \( x^2 \) (here \( a = -2 \)) determines the direction the parabola opens. Since \( a=-2<0 \), the parabola opens downward, which means the function has a maximum value (at the vertex) and no minimum value (it decreases without bound).

Step2: Determine the smallest y - value

Because the parabola opens downward, as \( x \) approaches \( \pm\infty \), \( y \) will approach \( -\infty \). So there is no smallest (minimum) value for this function; it decreases without bound.

Answer:

\(-\infty\)