Respuesta :
Answer:
The vertex for this parabola is [tex](\frac{3}{4}, -\frac{41}{4})[/tex]
The vertical intercept is y = -8
The solutions are [tex]x_{1}=\frac{3+\sqrt{41}}{4},\:x_{2}=\frac{3-\sqrt{41}}{4}[/tex]
Step-by-step explanation:
A) To find the vertex we need to follow this steps:
If we are given the general form of a quadratic function:
[tex]f(x) = ax^2+bx+c[/tex]
We can define the vertex, (h,k), by doing the following:
- Identify a, b, and c.
- Find the x-coordinate of the vertex by substituting a and b into [tex]-\frac{b}{2a}[/tex]
- Find the y-coordinate of the vertex by evaluating the x-coordinate into the quadratic function [tex]f(-\frac{b}{2a})[/tex]
We know that the equation is f(x) = 4x² - 6x - 8
a = 4, b = -6, and c = -8
The x-coordinate of the vertex is [tex]-\frac{-6}{2\cdot 4} = \frac{3}{4}[/tex]
The y-coordinate of the vertex is [tex]4(\frac{3}{4})^{2} - 6(\frac{3}{4}) - 8 = -\frac{41}{4}[/tex]
B) To find the vertical intercept we need to:
Evaluate x = 0 into the quadratic function given
[tex]y = 4(0)^{2} - 6(0) - 8 = -8[/tex]
C) To find the coordinates of the two intercepts of the parabola is the same as to find the solutions of the quadratic equation
For this we can use the Quadratic equation formula
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\mathrm{For\:}\quad a=4,\:b=-6,\:c=-8:\quad x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:4\left(-8\right)}}{2\cdot \:4}[/tex]
[tex]x_{1} =\frac{-\left(-6\right)+\sqrt{\left(-6\right)^2-4\cdot \:4\left(-8\right)}}{2\cdot \:4}= \frac{3+\sqrt{41}}{4}[/tex]
[tex]x_{2}=\frac{-\left(-6\right)-\sqrt{\left(-6\right)^2-4\cdot \:4\left(-8\right)}}{2\cdot \:4}=\frac{3-\sqrt{41}}{4}[/tex]
The solutions to the quadratic equation are:
[tex]x_{1}=\frac{3+\sqrt{41}}{4},\:x_{2}=\frac{3-\sqrt{41}}{4}[/tex]
