Question

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s). Determine the equation for the quadratic relationship graphed below.

Answer 1

Answer:

3x^2-6x-1

Step-by-step explanation:

f(x)=-1 has 2 roots x=0 and x=2 implies f(x)+1=0

f(x)+1 is quadratic since f(x).is quadratic

f(x)+1=kx(x-2) for some real k. f(1)=-4 implies k=3

Answer 2

The quadratic equation for the given parabola is $y = \frac{-(x)^2}{3} + (-\frac{8x}{3}) + (-\frac{13}{3}).$

Given:

From the graph, (h, k) = (-4, 1).

Let the point (0, -1) be taken as (x, y).

To Find:

We have to find the equation for the quadratic relationship graphed.

Solution:

The given graph represents a parabola with its vertex not at origin.

The vertex form of the equation for a parabola is given by,

$y = a(x - h)^2 + k$.

The standard form of the equation for a parabola is given by,

$y = ax^2+bx+c.$

On substituting the points (x, y) = (0, -1) and (h, k) = (-4, 1) in the vertex form of equation, we get,

$(-1) = a(0 - (-4))^2 + 1$

On simplifying, we get,

$(-1) = 16a + 1$

$i.e., a = \frac{-2}{6} = \frac{-1}{3}.$

From the graph, (h, k) = (-4, 1).

On substituting these values in the vertex form of equation, we get,

$y = a(x - (-4))^2 + 1$

On simplifying, the above equation becomes,

$y = a(x +4)^2 + 1$

Expanding $(x +4)^2$ using $(a + b)^2 = a^2 + 2ab +b^2$, we get,

$y = a(x^2 + 8x + 16) + 1$

Or, $y = ax^2 + 8ax + 16a + 1$.

Substituting the value of a in the above equation, we get,

$y = (\frac{-1}{3} )x^2 + 8.(\frac{-1}{3} ).x + 16.(\frac{-1}{3} ) + 1$

∴, $y = (\frac{-1}{3} )x^2 + (\frac{-8}{3} )x + (\frac{-13}{3} )$,

where $a = \frac{-1}{3}$, $b = \frac{-8}{3}$ and $c = \frac{-13}{3}$.

The above given is the standard equation for the given parabola.

Hence, the quadratic equation for the given parabola is $y = \frac{-(x)^2}{3} + (-\frac{8x}{3}) + (-\frac{13}{3}).$

#SPJ2