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1 year ago

use the graphing calculator to graph the quadratic function y = x2 − 5x − 36. Which value is a solution of 0 = x2 − 5x − 36?

Mathematics

1 answer:

daser333 [38]1 year ago

3 0

The solutions of the quadratic equation x² - 5x - 36 = 0 are -4 and 9.

The graph is attached below.

We are given a quadratic function.
A polynomial equation of degree two in one variable is a quadratic equation.
The function given to us is :
y = x² - 5x - 36
We need to find the solution of the quadratic function.
To find the roots, let y = 0.
x² - 5x - 36 = 0
Use the quadratic formula.
In elementary algebra, the quadratic formula is a formula that gives the solution(s) to a quadratic equation.
x = [-b±√b²-4ac]/2a
x = [-(-5) ± √25 - 4(1)(-36)]/2(1)
x = (5 ± √25 + 144)/2
x = (5 ± √169)/2
x = (5 ± 13)/2
x = 9 or x = -4

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Quadrilateral ABCD with vertices A(0, 6), B(-3, -6), C(-9, -6), and D(-12, -3): a) dilation with scale factor of 1/3 centered at

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a) The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively.

b) The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively.

<h3>How to perform transformations with points</h3>

a) A dillation centered at the origin is defined by following operation:

$P'(x,y) = k\cdot P(x,y)$ (1)

Where:

$P(x,y)$ - Original point
$P'(x,y)$ - Dilated point.

If we know that $k = \frac{1}{3}$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ , then the new points of the quadrilateral are:

$A'(x,y) = \frac{1}{3}\cdot (0,6)$

$A'(x,y) = (0, 2)$

$B'(x,y) = \frac{1}{3} \cdot (-3,-6)$

$B'(x,y) = (-1, -2)$

$C'(x,y) = \frac{1}{3}\cdot (-9,-6)$

$C'(x,y) = \left(-3,-2\right)$

$D'(x,y) = \frac{1}{3}\cdot (-12,-3)$

$D'(x,y) = (-4, -1)$

The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively. $\blacksquare$

b) A translation along a vector is defined by following operation:

$P'(x,y) = P(x,y) +T(x,y)$ (2)

Where $T(x,y)$ is the transformation vector.

If we know that $T(x,y) = (-5,-1)$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ ,

$A'(x,y) = (0,6) + (-5, -1)$

$A'(x,y) = (-5, 5)$

$B'(x,y) = (-3, -6) + (-5, -1)$

$B'(x,y) = (-8,-7)$

$C'(x,y) = (-9, -6) + (-5, -1)$

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$D'(x,y) = (-17, -4)$

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Vertex form: f (x) = a(x - h) 2 + k

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(h, k) is the vertex of the parabola-

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Feel free to ask for more if needed or if you did not understand something.

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