Question

1. $y = -x^2 + 2x + 3$

Explanation:

Step1: Find the vertex of the parabola

For a quadratic function \( y = ax^{2}+bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). In the function \( y=-x^{2}+2x + 3 \), \( a=- 1 \), \( b = 2 \), \( c = 3 \).
So \( x=-\frac{2}{2\times(-1)}=\frac{-2}{-2}=1 \)
Substitute \( x = 1 \) into the function to find the y - coordinate of the vertex:
\( y=-(1)^{2}+2\times1 + 3=-1 + 2+3=4 \)
So the vertex is \( (1,4) \)

Step2: Find the y - intercept

The y - intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function:
\( y=-(0)^{2}+2\times0+3 = 3 \)
So the y - intercept is \( (0,3) \)

Step3: Find the x - intercepts

Set \( y = 0 \), then \( -x^{2}+2x + 3=0 \)
Multiply both sides by - 1: \( x^{2}-2x - 3=0 \)
Factor the quadratic equation: \( x^{2}-2x - 3=(x - 3)(x+1)=0 \)
Set each factor equal to zero:
\( x - 3=0\Rightarrow x = 3 \)
\( x + 1=0\Rightarrow x=-1 \)
So the x - intercepts are \( (3,0) \) and \( (-1,0) \)

Step4: Plot the points and draw the parabola

We have the vertex \( (1,4) \), y - intercept \( (0,3) \), x - intercepts \( (3,0) \) and \( (-1,0) \). We can also find a few more points for accuracy. For example, when \( x = 2 \), \( y=-(2)^{2}+2\times2+3=-4 + 4+3 = 3 \), so the point \( (2,3) \) is on the parabola.
Now we can plot these points \( (1,4) \), \( (0,3) \), \( (3,0) \), \( (-1,0) \), \( (2,3) \) on the coordinate plane and draw a parabola opening downwards (since \( a=-1<0 \)) passing through these points.

Answer:

To graph \( y=-x^{2}+2x + 3 \):

• Vertex: \( (1,4) \)
• Y - intercept: \( (0,3) \)
• X - intercepts: \( (-1,0) \) and \( (3,0) \)
• Additional point: \( (2,3) \)

Plot these points and draw a downward - opening parabola through them.