Question

I need help with a rather long question in math and it’s been a challenge.

Answer 1

Step-by-step explanation

Since we have the given quadratic function:

`y=-4x^2-8x-8`

a)

Since the value of the squared term coefficient is negative, thus the graph opens up.

b)

`\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-(b)/(2a)`
`\mathrm{The\: parabola\: params\: are\colon}`
`a=-4,\: b=-8,\: c=-8`
`x_v=-(b)/(2a)`
`x_v=-(\left(-8\right))/(2\left(-4\right))`
`\mathrm{Simplify}`
`x_v=-1`

Plugging in the x_v value into the equation:

`y_v=-4\mleft(-1\mright)^2-8\mleft(-1\mright)-8`

Computing the powers:

`y_v=-4\cdot1-8(-1)-8`

Multiplying numbers:

`y_v=-4+8-8`

Adding numbers:

`y_v=-4`

`\mathrm{Therefore\: the\: parabola\: vertex\: is}`
`\mleft(-1,\: -4\mright)`

c) For a quadratic equation of the form ax^2 + bx + c = 0 the discriminant is b^2 -4ac

`\mathrm{For\: }\quad a=-4,\: b=-8,\: c=-8\colon\quad \mleft(-8\mright)^2-4\mleft(-4\mright)\mleft(-8\mright)`

Computing the powers and multiplying numbers:

`=-64`

Since the discriminant cannot be negative, there are not x-intercepts.

d) Identifying y-intercepts:

`y\mathrm{-intercept\: is\: the\: point\: on\: the\: graph\: where\: }x=0`
`y=-4\cdot\: 0^2-8\cdot\: 0-8`
`\mathrm{Apply\: rule}\: 0^a=0`
`y=-4\cdot\: 0-8\cdot\: 0-8`
`\mathrm{Subtract\: the\: numbers\colon}\: -0-0-8=-8`
`y=-8`

The y-intercept is at (0,-8)