QuestionConsider the parabola given by the equation: $$y=-1 x^{2}+2 x+3$$
Find the following for this parabola: A) The vertex $=($ $\square$ $\square$ ) B) The $y$ intercept is the point ( 0 , $\square$ ) C) Find the two values of $x$ that correspond to the $x$ intercepts of the parabola and write them as a list, separated by commas: $x=$ $\square$

Studdy Solution
Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the x-intercepts.
$$x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(3)}}{2(-1)}$$ $$x = \frac{-2 \pm \sqrt{4 + 12}}{-2}$$ $$x = \frac{-2 \pm \sqrt{16}}{-2}$$ $$x = \frac{-2 \pm 4}{-2}$$
Calculate the two solutions:
$$x_1 = \frac{-2 + 4}{-2} = -1$$ $$x_2 = \frac{-2 - 4}{-2} = 3$$
The x-intercepts are $x = -1, 3$.
A) The vertex is $(1, 4)$.
B) The y-intercept is the point $(0, 3)$.
C) The x-intercepts are $x = -1, 3$.