Questions: Enter the vertex: Enter the x-intercept (3): Enter the y-intercept(s): Choose the correct graph.

Transcript text: Enter the vertex: $\square$ Enter the x-intercept (3): $\square$ Enter the y-intercept(s): $\square$ Choose the correct graph.

Solution

Solution Steps

Step 1: Find the vertex.

The given quadratic function is $f(x) = -x^2 + 2x + 5$. The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$, where $a = -1$ and $b = 2$. So, $x = -\frac{2}{2(-1)} = 1$. The y-coordinate is $f(1) = -(1)^2 + 2(1) + 5 = -1 + 2 + 5 = 6$. Thus, the vertex is $(1, 6)$.

Step 2: Find the x-intercepts.

To find the x-intercepts, we set $f(x) = 0$ and solve for $x$. $-x^2 + 2x + 5 = 0$ $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(-1)(5)}}{2(-1)} = \frac{-2 \pm \sqrt{4 + 20}}{-2} = \frac{-2 \pm \sqrt{24}}{-2} = \frac{-2 \pm 2\sqrt{6}}{-2} = 1 \mp \sqrt{6}$ $x \approx 1 \pm 2.45$ $x_1 \approx 1 - 2.45 = -1.45$, $x_2 \approx 1 + 2.45 = 3.45$ The x-intercepts are approximately $(-1.5, 0)$ and $(3.5, 0)$.

Step 3: Find the y-intercept.

To find the y-intercept, we set $x = 0$ in the function $f(x)$: $f(0) = -(0)^2 + 2(0) + 5 = 5$ The y-intercept is $(0, 5)$.