Question

question
find the minimum value of the function $f(x) = 1.4x^2 + 13x + 24$ to the nearest hundredth.

Explanation:

Step1: Identify the coefficients

For the quadratic function \( f(x) = ax^2 + bx + c \), here \( a = 1.4 \), \( b = 13 \), \( c = 24 \).

Step2: Find the x-coordinate of the vertex

The formula for the x-coordinate of the vertex of a quadratic function is \( x = -\frac{b}{2a} \). Substituting the values, we get \( x = -\frac{13}{2\times1.4} = -\frac{13}{2.8} \approx -4.6429 \).

Step3: Find the minimum value (y-coordinate of the vertex)

Substitute \( x \approx -4.6429 \) into the function \( f(x) \):
\( f(-4.6429) = 1.4\times(-4.6429)^2 + 13\times(-4.6429) + 24 \)
First, calculate \( (-4.6429)^2 \approx 21.5566 \)
Then, \( 1.4\times21.5566 \approx 30.1792 \)
Next, \( 13\times(-4.6429) \approx -60.3577 \)
Now, add them up: \( 30.1792 - 60.3577 + 24 \approx -6.1785 \approx -6.18 \) (to the nearest hundredth)

Answer:

\( -6.18 \)