Question

graph the quadratic function (f(x)=x^{2}+2x - 3). *to create the graph of the parabola: (1) select parabola (2) plot the vertex (3) plot the y - intercept.

Explanation:

Step1: Find the vertex

For a quadratic function $y = ax^{2}+bx + c$ (here $a = 1$, $b = 2$, $c=-3$), the $x$-coordinate of the vertex is $x=-\frac{b}{2a}$. So $x =-\frac{2}{2\times1}=- 1$. Substitute $x = - 1$ into $y=x^{2}+2x - 3$, we get $y=(-1)^{2}+2\times(-1)-3=1 - 2 - 3=-4$. The vertex is $(-1,-4)$.

Step2: Find the y - intercept

Set $x = 0$ in the function $y=x^{2}+2x - 3$. Then $y=0^{2}+2\times0 - 3=-3$. The y - intercept is $(0,-3)$.

Step3: Sketch the parabola

Since $a = 1>0$, the parabola opens upwards. Plot the vertex $(-1,-4)$ and the y - intercept $(0,-3)$ and then draw a smooth curve to form the parabola.

Answer:

Graph a parabola opening upwards with vertex at $(-1,-4)$ and y - intercept at $(0,-3)$.