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Marta_Voda [28]

3 years ago

6

Find the axis of symmetry y= -2x^2 -8x -15

1 answer:

katen-ka-za [31]3 years ago

4 0

The squared variable is the "x", therefore is a vertical parabola so its axis of symmetry will come from the vertex's x-coordinate.

now, let's find the vertex of it.

$\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
y=\stackrel{\stackrel{a}{\downarrow }}{-2}x^2\stackrel{\stackrel{b}{\downarrow }}{-8}x\stackrel{\stackrel{c}{\downarrow }}{-15}
\qquad \qquad 
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left( -\cfrac{-8}{2(-2)}~~,~~\qquad \quad \right)\implies \left( -2~~,~~\qquad \right)
\\\\\\
\stackrel{\textit{axis of symmetry}}{x=-2}$

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A jumping spider's movement is modeled by a parabola. The spider makes a single jump from the origin and reaches a maximum heigh

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A parabola is a mirror-symmetrical U-shape.

The equation of the parabola is $\mathbf{y = -\frac{1}{640}(x - 80)^2 + 10}$
The focus is $\mathbf{Focus = (80, -1760)}$
The directrix is $\mathbf{y = \frac{1}{640}}$
The axis of the symmetry of parabola is: $\mathbf{x = 80}$

From the question, we have:

$\mathbf{Vertex: (h,k) = (80,10)}$

$\mathbf{Origin: (x,y) = (0,0)}$

The equation of a parabola is:

$\mathbf{y = a(x - h)^2 + k}$

Substitute the values of origin and vertex in $\mathbf{y = a(x - h)^2 + k}$

$\mathbf{0 = a(0 - 80)^2 + 10}$

$\mathbf{0 = a(- 80)^2 + 10}$

$\mathbf{0 = 6400a + 10}$

Collect like terms

$\mathbf{6400a =- 10}$

Solve for a

$\mathbf{a =- \frac{1}{640}}$

Substitute the values of a and the vertex in $\mathbf{y = a(x - h)^2 + k}$

$\mathbf{y = -\frac{1}{640}(x - 80)^2 + 10}$

The focus of a parabola is:

$\mathbf{Focus = (h, \frac{k+1}{4a})}$

Substitute the values of a and the vertex in $\mathbf{Focus = (h, \frac{k+1}{4a})}$

$\mathbf{Focus = (80, \frac{10+1}{4 \times -\frac{1}{640}})}$

$\mathbf{Focus = (80, -\frac{11}{\frac{1}{160}})}$

$\mathbf{Focus = (80, -11\times 160)}$

$\mathbf{Focus = (80, -1760)}$

The equation of the directrix is:

$\mathbf{y = -a}$

So, we have:

$\mathbf{y = \frac{1}{640}}$ ----- the directrix

The axis of symmetry is:

$\mathbf{x = -\frac{b}{2a}}$

We have:

$\mathbf{y = -\frac{1}{640}(x - 80)^2 + 10}$

Expand

$\mathbf{y = -\frac{1}{640}(x^2 -160x + 6400) +10}$

Expand

$\mathbf{y = -\frac{1}{640}x^2 +\frac{1}{4}x - 10 +10}$

$\mathbf{y = -\frac{1}{640}x^2 +\frac{1}{4}x }$

A quadratic function is represented as:

$\mathbf{y = ax^2 + bx + c}$

So, we have:

$\mathbf{a = -\frac{1}{640}}$

$\mathbf{b = \frac{1}{4}}$

Recall that:

$\mathbf{x = -\frac{b}{2a}}$

So, we have:

$\mathbf{x = -\frac{1/4}{2 \times -1/640}}$

$\mathbf{x = \frac{1/4}{1/320}}$

This gives

$\mathbf{x = \frac{320}{4}}$

$\mathbf{x = 80}$

Hence, the axis of the symmetry of parabola is: $\mathbf{x = 80}$

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