instructions Consider the function f(x)=x^2+bx-16 where b is a constant. Activity 1 of 2 If the function has an axis of symmetry at x=5 what is the value of b? b=-10 Activity 2 of 2 If b=-6 what are the zero(s ) of the function? x_(1)=? Choose. and x_(2)= square v Choose -1 -6 -8 -2

To find the value of \( b \) when the axis of symmetry is at \( x = 5 \), we can use the formula for the axis of symmetry of a quadratic function, which is given by:<br /><br />\[ x = -\frac{b}{2a} \]<br /><br />For the given function \( f(x) = x^2 + bx - 16 \), the coefficient \( a \) is 1. Plugging in the given axis of symmetry:<br /><br />\[ 5 = -\frac{b}{2 \cdot 1} \]<br /><br />Solving for \( b \):<br /><br />\[ 5 = -\frac{b}{2} \]<br /><br />\[ 10 = -b \]<br /><br />\[ b = -10 \]<br /><br />So, the value of \( b \) is \( -10 \).<br /><br />Now, let's find the zeros of the function when \( b = -6 \). The function becomes:<br /><br />\[ f(x) = x^2 - 6x - 16 \]<br /><br />To find the zeros, we solve the equation:<br /><br />\[ x^2 - 6x - 16 = 0 \]<br /><br />We can use the quadratic formula:<br /><br />\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]<br /><br />For our equation, \( a = 1 \), \( b = -6 \), and \( c = -16 \). Plugging these values into the quadratic formula:<br /><br />\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \]<br /><br />\[ x = \frac{6 \pm \sqrt{36 + 64}}{2} \]<br /><br />\[ x = \frac{6 \pm \sqrt{100}}{2} \]<br /><br />\[ x = \frac{6 \pm 10}{2} \]<br /><br />This gives us two solutions:<br /><br />\[ x_1 = \frac{6 + 10}{2} = 8 \]<br /><br />\[ x_2 = \frac{6 - 10}{2} = -2 \]<br /><br />So, the zeros of the function are \( x_1 = 8 \) and \( x_2 = -2 \).<br /><br />Therefore, the correct choices are:<br /><br />\[ x_1 = 8 \quad \text{and} \quad x_2 = -2 \]