Question

What is the vertex form of y=-2x^2-8x+10? Give the coordinates of the vertex, state if there is a minimum or maximum, and find the equation of the line of symmetry.

This is for Algebra 2 regular.

Answer 1

first off, let's notice that the squared variable is the "x", that means is a vertically opening parabola.

notice the leading term's coefficient, is -2, is negative, that means the parabola is opening downwards, is facing down, so it goes up up up, reaches the U-turn and then down down down, is a "hump" or a maximum point.


`\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ \begin{array}{lcccl} y = & -2x^2& -8x& +10\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{-8}{2(-2)}~~,~~10-\cfrac{(-8)^2}{4(-2)} \right)\implies \left( \cfrac{8}{-4}~~,~~10-\cfrac{64}{-8} \right) \\\\\\ \left( -2~~,~~10+8 \right)\implies (-2~~,~~18)`

and the line of symmetry, well is a vertical parabola, mirroring itself at the line x = -2, which is the x-coordinate of the vertex.