To find the x-intercept(s), we can use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
This formula is applied when we have a polynomial of the form:
[tex]ax^2+bx+c=0[/tex]
We have that our case is:
[tex]\frac{1}{4}x^2+\frac{3}{2}x-\frac{1}{8}=0[/tex]
To find the x-intercepts of the parabola. However, we can find the least common multiple of the denominators to have an easier equation to solve. Then, we have:
[tex]\operatorname{lcm}(4,2,8)=8[/tex]
Then, we can multiply each side of the equation by 8 as follows:
[tex]8(\frac{1}{4}x^2+\frac{3}{2}x-\frac{1}{8}=0)[/tex]
[tex]\frac{8}{4}x^2+\frac{8\cdot3}{2}x-\frac{8\cdot1}{8}=8\cdot0\Rightarrow2x^2+12x-1=0[/tex]
Now, we have that:
a = 2
b = 12
c = -1
Then, we have:
[tex]x=\frac{-12\pm\sqrt[]{12^2^{}-4(2)(-1)}}{2\cdot2}[/tex]
[tex]x=\frac{-12\pm\sqrt[]{144+8}}{4}\Rightarrow x=\frac{-12\pm\sqrt[]{152}}{4}[/tex]
Then
[tex]\sqrt[]{152}=\sqrt[]{2^2\cdot2\cdot19}=2\cdot\sqrt[]{38}[/tex]
Then, the two x-intercepts are:
[tex]x=\frac{-12\pm2\cdot\sqrt[]{38}}{4}[/tex]
[tex]x=\frac{-12+2\cdot\sqrt[]{38}}{4}\Rightarrow x=-\frac{12}{4}+\frac{2\cdot\sqrt[]{38}}{4}=-3+\frac{\sqrt[]{38}}{2}[/tex]
And
[tex]x=\frac{-12-2\cdot\sqrt[]{38}}{4}\Rightarrow x=-\frac{12}{4}-\frac{2\cdot\sqrt[]{38}}{4}=-3-\frac{\sqrt[]{38}}{2}[/tex]
Therefore:
The smallest x-intercept is:
[tex]-3-\frac{\sqrt[]{38}}{2}[/tex]
The largest x-intercept is:
[tex]-3+\frac{\sqrt[]{38}}{2}[/tex]
Finding the value of the vertex
To find the vertex of the parabola, we can find its x-value, and y-value using the following formulas:
[tex]x_v=-\frac{b}{2a},y_v=c-\frac{b^2}{4a}_{}[/tex]
In the original equation, we have:
a = 1/4
b = 3/2
c = -1/8
Then, we have:
[tex]x_v=-\frac{\frac{3}{2}}{2(\frac{1}{4})_{}}=-\frac{\frac{3}{2}}{\frac{2}{4}}=-\frac{3}{2}\frac{4}{2}=-\frac{3\cdot4}{4}\Rightarrow x_v=-3[/tex]
And
[tex]y_v=-\frac{1}{8}-\frac{(\frac{3}{2})^2}{4(\frac{1}{4})}=-\frac{1}{8}-\frac{\frac{9}{4}}{1}=-\frac{1}{8}-\frac{9}{4}=-(\frac{1}{8}+\frac{9}{4})=-(\frac{4+72}{32})[/tex]
Finally:
[tex]y_v=-\frac{76}{32}\Rightarrow y_v=-\frac{19}{8}[/tex]
Therefore, the vertex of the parabola is:
[tex](-3,-\frac{19}{8})[/tex]
In summary, we have:
The coordinates for the vertex are:
[tex](-3,-\frac{19}{8})[/tex]
The smallest x-intercept (only the
