Question

open-ended question
describe what you know about the
product $x^2 - 4x - 21$

Explanation:

Step1: Factor the quadratic expression

To factor \(x^{2}-4x - 21\), we need to find two numbers that multiply to \(- 21\) and add up to \(-4\). The numbers are \(-7\) and \(3\) since \(-7\times3=-21\) and \(-7 + 3=-4\).
So, \(x^{2}-4x - 21=(x - 7)(x+3)\)

Step2: Analyze the roots

Set the factored form equal to zero: \((x - 7)(x + 3)=0\)
Using the zero - product property, we get \(x-7 = 0\) or \(x + 3=0\)
Solving for \(x\), we have \(x = 7\) or \(x=-3\). These are the roots of the quadratic equation \(x^{2}-4x - 21 = 0\)

Step3: Analyze the graph of the quadratic function

The quadratic function \(y=x^{2}-4x - 21\) is a parabola. Since the coefficient of \(x^{2}\) is positive (\(a = 1>0\)), the parabola opens upwards.
The vertex of the parabola \(y=ax^{2}+bx + c\) has its \(x\) - coordinate given by \(x=-\frac{b}{2a}\). For \(y=x^{2}-4x - 21\), \(a = 1\) and \(b=-4\), so \(x=-\frac{-4}{2\times1}=2\)
Substitute \(x = 2\) into the function: \(y=(2)^{2}-4\times2-21=4-8 - 21=-25\). So the vertex is at \((2,-25)\), which is the minimum point of the parabola.

Answer:

The expression \(x^{2}-4x - 21\) is a quadratic trinomial. It can be factored as \((x - 7)(x + 3)\). The roots of the quadratic equation \(x^{2}-4x - 21=0\) are \(x = 7\) and \(x=-3\). The graph of the quadratic function \(y=x^{2}-4x - 21\) is a parabola opening upwards with vertex at \((2,-25)\).