x + 3 f ( x )

We are given the quadratic function:

$f(x) = x^2 + x + 3$

Let's go step by step to find the requested characteristics:

a) Vertex (as an ordered pair)

The vertex of a quadratic function $f(x) = ax^2 + bx + c$ can be found using the formula:

$x = \frac{-b}{2a}$

In this case, $a = 1$, $b = 1$, and $c = 3$.

$x = \frac{-1}{2(1)} = \frac{-1}{2}$

Now, plug this $x$-value into the function to find the $y$-coordinate of the vertex:

$f\left(\frac{-1}{2}\right) = \left(\frac{-1}{2}\right)^2 + \frac{-1}{2} + 3 = \frac{1}{4} - \frac{1}{2} + 3 = \frac{1}{4} - \frac{2}{4} + 3 = \frac{-1}{4} + 3 = \frac{11}{4}$

Thus, the vertex is:

$\left( \frac{-1}{2}, \frac{11}{4} \right)$

b) Equation of the line of symmetry

The axis of symmetry is the vertical line passing through the vertex, so its equation is:

$x = \frac{-1}{2}$

c) Range (in interval notation)

Since this is a parabola that opens upwards (as $a = 1 > 0$), the range starts at the $y$-coordinate of the vertex and goes to infinity. The $y$-coordinate of the vertex is $\frac{11}{4}$.

Thus, the range is:

$\left[ \frac{11}{4}, \infty \right)$

d) $x$-intercept(s) (as ordered pair(s))

To find the $x$-intercepts, set $f(x) = 0$ and solve for $x$:

$x^2 + x + 3 = 0$

Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = 1$, and $c = 3$. Plugging these into the formula:

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(3)}}{2(1)} = \frac{-1 \pm \sqrt{1 - 12}}{2} = \frac{-1 \pm \sqrt{-11}}{2}$

Since the discriminant is negative ($\sqrt{-11}$), there are no real $x$-intercepts. Thus, the answer is:

$\text{DNE}$

e) $y$-intercept (as ordered pair)

To find the $y$-intercept, set $x = 0$ and solve for $y$:

$f(0) = 0^2 + 0 + 3 = 3$

Thus, the $y$-intercept is:

$(0, 3)$

f) Graph of the parabola

The parabola opens upwards, with a vertex at $\left( \frac{-1}{2}, \frac{11}{4} \right)$, no $x$-intercepts, and a $y$-intercept at $(0, 3)$. The graph should show a parabola opening upwards, with the vertex above the x-axis.

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Final Answers:

• a) Vertex: $\left( \frac{-1}{2}, \frac{11}{4} \right)$
• b) Equation of the line of symmetry: $x = \frac{-1}{2}$
• c) Range: $\left[ \frac{11}{4}, \infty \right)$
• d) $x$-intercept(s): DNE
• e) $y$-intercept: $(0, 3)$

Would you like more details on any part of the process?

Related Questions:

1. What happens to the vertex if we change the constant term?
2. How does the discriminant of the quadratic formula relate to the number of real intercepts?
3. What does the range of a quadratic function depend on?
4. How would the graph look if the leading coefficient were negative?
5. How can we graph a quadratic function using the vertex and intercepts?

Tip:

The quadratic formula works for all quadratic equations, even if the discriminant is negative—it just means the roots are complex!