Answer:
See explanation
Step-by-step explanation:
Given the quadratic function [tex]y=2x^2+8x-5[/tex]
To plot the graph of this function, find the vertex of parabola, x- and y- intercepts.
1. The vertex:
[tex]x_v=\dfrac{-b}{2a}=\dfrac{-8}{2\cdot 2}=-\dfrac{8}{4}=-2\\ \\y_v=2\cdot (-2)^2+8\cdot (-2)-5=8-16-5=-13[/tex]
2. y-intercept:
[tex]x=0\\ \\y=2\cdot 0^2+8\cdot 0-5=-5[/tex]
3. x-intercepts:
[tex]y=0\\ \\2x^2+8x-5=0\\ \\D=b^2-4ac=8^2-4\cdot 2\cdot (-5)=64+40=104\\ \\\sqrt{D}=\sqrt{104}\approx 10.1980\\ \\x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{-8\pm 10.1980}{2\cdot 2}\approx -4.55,\ 0.55[/tex]
4. The leading coefficient is 2 > 0, then parabola goes in positive y-direction
