Question

section 05.1: problem 11 (4 points) results for this submission 1 of the answers is not correct. for the quadratic function f(x)=x² + 4x - 6, find the vertex and the x - intercept(s). express the vertex as a coordinate pair. if there is more than one intercept, enter as a comma separated list of points. vertex (-2,-10) help (points) x - intercept(s) (-2 + √10,0),(-2 - √10,0) help (points) note: you can earn partial credit on this problem. note: you are in the reduced scoring period. all work counts for 70% of the original. preview my answers submit answers show me another your score was recorded. scores are sent to d2l, brightspace every 24 hours. you have attempted this problem 3 times. you received a score of 35% for this attempt. your overall recorded score is 35%. you have unlimited attempts remaining.

Explanation:

Step1: Recall vertex - formula for quadratic function

For a quadratic function \(y = ax^{2}+bx + c\), the \(x\) - coordinate of the vertex is \(x=-\frac{b}{2a}\). In the function \(f(x)=x^{2}+4x - 6\), where \(a = 1\), \(b = 4\), and \(c=-6\). Then \(x=-\frac{4}{2\times1}=-2\).

Step2: Find the \(y\) - coordinate of the vertex

Substitute \(x = - 2\) into the function \(f(x)\): \(f(-2)=(-2)^{2}+4\times(-2)-6=4 - 8 - 6=-10\). So the vertex is \((-2,-10)\).

Step3: Find the \(x\) - intercepts

Set \(y = f(x)=0\), so \(x^{2}+4x - 6=0\). Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 1\), \(b = 4\), \(c=-6\). Then \(x=\frac{-4\pm\sqrt{4^{2}-4\times1\times(-6)}}{2\times1}=\frac{-4\pm\sqrt{16 + 24}}{2}=\frac{-4\pm\sqrt{40}}{2}=\frac{-4\pm2\sqrt{10}}{2}=-2\pm\sqrt{10}\). So the \(x\) - intercepts are \((-2+\sqrt{10},0)\) and \((-2 - \sqrt{10},0)\).

Answer:

vertex: \((-2,-10)\), \(x\) - intercepts: \((-2+\sqrt{10},0),(-2 - \sqrt{10},0)\)