Graph the quadratic function y = -x2+4x+1. Find the vertex, y-intercept, and x-intercepts.

The given equation $y = -x^2+4x+1$ is a quadratic function of the form $y = ax^2+bx+c$. Since the coefficient of $x^2$ is $a=-1$ (which is negative), the graph is a parabola that opens downwards.

The x-coordinate of the vertex is given by the formula $x = -\frac{b}{2a}$.
For $y = -x^2+4x+1$, we have $a=-1$ and $b=4$.

Substitute $x=2$ into the equation to find the y-coordinate of the vertex:

The vertex is at $(2, 5)$. This is the maximum point of the parabola.

The y-intercept occurs when $x=0$.

The y-intercept is at $(0, 1)$.

The x-intercepts occur when $y=0$.

Multiply by $-1$ to make the leading coefficient positive:

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:

The x-intercepts are approximately $x_1 = 2 - \sqrt{5} \approx 2 - 2.236 = -0.236$ and $x_2 = 2 + \sqrt{5} \approx 2 + 2.236 = 4.236$.
So, the x-intercepts are approximately $(-0.24, 0)$ and $(4.24, 0)$.

To sketch the graph, plot the following points:

• Vertex: $(2, 5)$
• Y-intercept: $(0, 1)$
• X-intercepts: $(2-\sqrt{5}, 0)$ and $(2+\sqrt{5}, 0)$ The parabola opens downwards, is symmetric about the vertical line $x=2$ (the axis of symmetry), and passes through these points.