Question 1209789: Draw the graph of y=x²+x+2
For -4≤x≤4
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! To draw the graph of $y = x^2 + x + 2$, we need to determine its key features:
1. **Vertex:** The vertex of the parabola is at the point $x = \dfrac{-b}{2a} = \dfrac{-1}{2(1)} = -\dfrac{1}{2}$. Substituting this value of $x$ into the equation, we get $y = \left( -\dfrac{1}{2} \right)^2 - \dfrac{1}{2} + 2 = \dfrac{7}{4}$. So, the vertex is at $\left( -\dfrac{1}{2}, \dfrac{7}{4} \right)$.
2. **Axis of Symmetry:** The axis of symmetry is the vertical line passing through the vertex, which is $x = -\dfrac{1}{2}$.
3. **y-intercept:** The y-intercept is the point where the graph intersects the y-axis. Setting $x = 0$, we get $y = 2$. So, the y-intercept is at $(0,2)$.
4. **x-intercepts:** The x-intercepts are the points where the graph intersects the x-axis. Setting $y = 0$, we get the quadratic equation $x^2 + x + 2 = 0$. The discriminant of this quadratic is $1^2 - 4 \cdot 1 \cdot 2 = -7$, which is negative, so the quadratic has no real roots. This means the graph does not intersect the x-axis.
5. **Direction:** Since the coefficient of $x^2$ is positive, the parabola opens upward.
Now, we can plot the graph:
[asy]
unitsize(2 cm);
real func (real x) {
return(x^2 + x + 2);
}
draw(graph(func,-4,4));
draw((-4,0)--(4,0));
draw((0,-1)--(0,18));
label("$y = x^2 + x + 2$", (4,func(4)), E);
dot("$\left( -\frac{1}{2}, \frac{7}{4} \right)$", (-1/2,7/4), S);
dot("$y$-intercept", (0,2), NE);
[/asy]
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