6. A function of the form \( p(x)=a x^{2}+q \) has two \( x \)-intercepts, one of which is Determine the other \( x \)-intercept and explain how you arrived at your answer. \( p(x)=a x^{2}+q \) \( 0=a x^{2}+q \) \( -q=a x^{\overline{2}} 9 \)

To find the other \( x \)-intercept of the function \( p(x) = ax^2 + q \), we start with the equation for the \( x \)-intercepts, which is \( 0 = ax^2 + q \). Rearranging gives us \( ax^2 = -q \). Since we are given that one of the \( x \)-intercepts is known (let's call it \( r_1 \)), we can substitute \( r_1 \) back into the equation to find \( q \). Thus, \( 0 = ar_1^2 + q \) implies \( q = -ar_1^2 \). Now, substituting this back into \( 0 = ax^2 + q\): \[ 0 = ax^2 - ar_1^2 \] Factoring gives us \( a(x^2 - r_1^2) = 0 \), which can be further factored as \( a(x - r_1)(x + r_1) = 0 \). Therefore, the other \( x \)-intercept is \( -r_1 \). So if you have one intercept at \( r_1 = 3 \), the other intercept would be \( -3 \). This neat symmetry in quadratic functions is a delightful feature of their parabolic nature! They always offer two \( x \)-intercepts that are equidistant from the vertex along the \( x \)-axis when the function has real roots.