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Need help with finding the vertex and the Y intercept of the quadratic function and use them to graph the function for number 21

Need Help With Finding The Vertex And The Y Intercept Of The Quadratic Function And Use Them To Graph The Function For Number 21 class=

Sagot :

EXPLANATION

Given the function y=-2x^2 -12x -5

[tex]\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-\frac{b}{2a}[/tex][tex]\mathrm{The\: parabola\: params\: are\colon}[/tex][tex]a=-2,\: b=-12,\: c=-5[/tex][tex]x_v=-\frac{b}{2a}[/tex][tex]x_v=-\frac{\left(-12\right)}{2\left(-2\right)}[/tex][tex]\mathrm{Simplify}\colon[/tex][tex]x_v=-3[/tex]

Plug in x_v=-3 to find the y_v value:

[tex]y_v=-2\mleft(-3\mright)^2-12\mleft(-3\mright)-5[/tex]

Computing the powers and multiplying terms:

[tex]y_v=-18+36-5[/tex]

Adding and subtracting numbers:

[tex]y_{v\text{ }}=13[/tex]

Therefore, the parabola vertex is:

(-3,13)

Now, we need to compute the y-intercept

[tex]y\mathrm{-intercept\: is\: the\: point\: on\: the\: graph\: where\: }x=0[/tex][tex]\mathrm{Apply\: rule}\: 0^a=0[/tex][tex]0^2=0[/tex][tex]y=-2\cdot\: 0-12\cdot\: 0-5[/tex]

Multiplying numbers:

[tex]y=-5[/tex]

The y-intercept is at (0,-5)

In conclusion, the graph of the function is as follows:

View image AmadeusI567469