Use the vertex and intercepts to sketch the graph of the quadratic function f(x) = x2 + 6x + 3. Give the equation of the parabola’s axis of symmetry. Use the graph to determine the function’s domain and range.

Answered: Use the vertex and intercepts to sketch…

College Algebra

10th Edition

ISBN:9781337282291

Author:Ron Larson

Publisher:Ron Larson

Chapter3: Polynomial Functions

Section3.1: Quadratic Functions And Models

Problem 4ECP: Write the standard form of the quadratic function whose graph is a parabola with vertex (4,11) and...

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Use the vertex and intercepts to sketch the graph of the quadratic function f(x) = x² + 6x + 3. Give the equation of the parabola’s axis of symmetry. Use the graph to determine the function’s domain and range.

Expert Solution

Step 1

Given, f(x) = x² + 6x + 3

Step 2

We know that when an equation is of the form Ax² + Bx + C then the vertex is $(- \frac{B}{2 A}, f (- \frac{B}{2 A}))$

Here A=1, B=6

To find vertex

$x = - \frac{6}{2 \times 1} = - 3 y = f (- 3) = {(- 3)}^{2} + 6 (- 3) + 3 = 9 - 18 + 3 = - 6$

Hence vertex is (-3,-6).

To find y-intercepts set x=0 we get.

y=3

Hence y-intercept is (0,3)

To find x-intercepts set y=0 and solve for x.

$\begin{array}{rcl} x^{2} + 6 x + 3 & = & 0 \\ \Rightarrow & x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{- 6 \pm \sqrt{6^{2} - 4 \times 1 \times 3}}{2 \times 1} \\ = & \frac{- 6 \pm \sqrt{36 - 12}}{2} \\ = & \frac{- 6 \pm \sqrt{24}}{2} \\ = & \frac{- 6 \pm \sqrt{4 \times 6}}{2} \\ = & \frac{- 6 \pm 2 \sqrt{6}}{2} \\ = & - 3 \pm \sqrt{6} \\ = & - 3 + \sqrt{6}, - 3 - \sqrt{6} \end{array}$

Hence x-intercepts are $(- 3 + \sqrt{6}, 0), (- 3 - \sqrt{6}, 0)$

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