Question

complete the sentence about the quadratic function ( y = -8x^2 + 16x + 1 ).
the domain of this function is (\boldsymbol{\text{all real numbers}}), and the range is (\boldsymbol{\text{{x | x ≥ -8}, {x | x ≥ 1}, {y | y ≥ 1}, {y | y ≤ 9}}}).

Explanation:

Step1: Analyze the domain of quadratic function

A quadratic function in the form \(y = ax^2+bx + c\) (here \(a=- 8\), \(b = 16\), \(c = 1\)) has a domain of all real numbers because there are no restrictions on the values of \(x\) (we can plug in any real number for \(x\) and get a valid \(y\) - value).

Step2: Analyze the range of the quadratic function

First, we can rewrite the quadratic function in vertex form. The formula for the \(x\) - coordinate of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is \(x=-\frac{b}{2a}\). For \(y=-8x^{2}+16x + 1\), \(a=-8\), \(b = 16\), so \(x=-\frac{16}{2\times(-8)}=\frac{- 16}{-16}=1\).
Now we find the \(y\) - coordinate of the vertex by plugging \(x = 1\) into the function: \(y=-8(1)^{2}+16(1)+1=-8 + 16+1=9\).
Since \(a=-8<0\), the parabola opens downwards. So the maximum value of the function is \(y = 9\) and the range is all real numbers \(y\) such that \(y\leqslant9\).

Answer:

The domain of this function is all real numbers, and the range is \(\{y|y\leqslant9\}\)