Sketch the graph of the quadratic function y = -x2+4x+1, identifying its vertex, y-intercept, and x-intercepts.

The coefficient of $x^2$ is $-1$, which is negative. Therefore, the parabola opens downwards.

For a quadratic function $y = ax^2+bx+c$, the x-coordinate of the vertex is given by $x = -\frac{b}{2a}$.
Here, $a=-1$, $b=4$, $c=1$.

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

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 $-1$:

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

The x-intercepts are approximately $(2+\sqrt{5}, 0) \approx (4.24, 0)$ and $(2-\sqrt{5}, 0) \approx (-0.24, 0)$.

• Plot the vertex $(2, 5)$.
• Plot the y-intercept $(0, 1)$.
• Plot the x-intercepts $(2+\sqrt{5}, 0)$ and $(2-\sqrt{5}, 0)$.
• Draw a smooth parabola opening downwards, passing through these points. The axis of symmetry is the vertical line $x=2$.

The key points for the sketch are:

• Vertex: $\boxed{\mathbf{(2, 5)}}$
• Y-intercept: $\boxed{\mathbf{(0, 1)}}$
• X-intercepts: $\boxed{\mathbf{(2-\sqrt{5}, 0) \text{ and } (2+\sqrt{5}, 0)}}$