Respuesta :
Answer:
The vertex: [tex](\frac{3}{4},-\frac{41}{4} )[/tex]
The vertical intercept is: [tex]y=-8[/tex]
The coordinates of the two intercepts of the parabola are [tex](\frac{3+\sqrt{41} }{4} , 0)[/tex] and [tex](\frac{3-\sqrt{41} }{4} , 0)[/tex]
Step-by-step explanation:
To find the vertex of the parabola [tex]4x^2-6x-8[/tex] you need to:
1. Find the coefficients a, b, and c of the parabola equation
[tex]a=4, b=-6, \:and \:c=-8[/tex]
2. You can apply this formula to find x-coordinate of the vertex
[tex]x=-\frac{b}{2a}[/tex], so
[tex]x=-\frac{-6}{2\cdot 4}\\x=\frac{3}{4}[/tex]
3. To find the y-coordinate of the vertex you use the parabola equation and x-coordinate of the vertex ([tex]f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c[/tex])
[tex]f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\f(\frac{3}{4})=4\cdot (\frac{3}{4})^2-6\cdot (\frac{3}{4})-8\\y=\frac{-41}{4}[/tex]
To find the vertical intercept you need to evaluate x = 0 into the parabola equation
[tex]f(x)=4x^2-6x-8\\f(0)=4(0)^2-6\cdot 0-0\\f(0)=-8[/tex]
To find the coordinates of the two intercepts of the parabola you need to solve the parabola by completing the square
[tex]\mathrm{Add\:}8\mathrm{\:to\:both\:sides}[/tex]
[tex]x^2-6x-8+8=0+8[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]4x^2-6x=8[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}4[/tex]
[tex]\frac{4x^2-6x}{4}=\frac{8}{4}\\x^2-\frac{3x}{2}=2[/tex]
[tex]\mathrm{Write\:equation\:in\:the\:form:\:\:}x^2+2ax+a^2=\left(x+a\right)^2[/tex]
[tex]x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=2+\left(-\frac{3}{4}\right)^2\\x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{41}{16}[/tex]
[tex]\left(x-\frac{3}{4}\right)^2=\frac{41}{16}[/tex]
[tex]\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}[/tex]
[tex]x_1=\frac{\sqrt{41}+3}{4},\:x_2=\frac{-\sqrt{41}+3}{4}[/tex]
