Use your calculator or other handheld technology to graph \( y=4 x^{2}+100 \). Then answer the following questions. What is the vertex of the graph? Give your answer in coordinate form. The vertex is 0,100 Are there any \( x \)-intercepts? ? Yes No (

The vertex of the graph \( y=4x^{2}+100 \) is located at the point (0, 100). In fact, since this is a quadratic function in the standard form \( y = ax^2 + bx + c \), and given that \( a = 4 \) (which is positive), the parabola opens upwards, meaning the vertex is indeed the minimum point of the graph. As for the \( x \)-intercepts, let's solve for when \( y = 0 \) in the equation \( 4x^{2} + 100 = 0 \). Rearranging gives us \( 4x^{2} = -100 \). Since we can't have a negative value on the left (as \( 4x^{2} \) is always non-negative for real x), we find that there are no real \( x \)-intercepts for this graph!