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2 years ago

Giving brainliest !!!!!!!!!!!

Mathematics

1 answer:

oksano4ka [1.4K]2 years ago

3 0

The parabola vertex is (1,5), the focus of the parabola is (1,6), and the directrix y = 4.

<h3>What is the graph of a parabolic equation?</h3>

The graph of a parabolic equation is a U-shape curve graph that is established from a quadratic equation.

From the given information:

$\mathbf{y=\dfrac{1}{4}(x-1)^2+5}$

The vertex of an up-down facing parabolic equation takes the form:

y = ax² + bx + c is $\mathbf{x_v = -\dfrac{b}{2a}}$

Rewriting the given equation:

$\mathbf{y = \dfrac{x^2}{4}-\dfrac{x}{2}+\dfrac{21}{4}}$

$\mathbf{x_v = -\dfrac{b}{2a}}$

$\mathbf{x_v = -\dfrac{(-\dfrac{1}{2})}{2(\dfrac{1}{4})}}$

$\mathbf{x_v =1}$

Replacing the value of x into the equation, y becomes:

$\mathbf{y_v = 5}$

Thus, the parabola vertex is (1,5)

From the vertex, the focus of the parabola is (1,6), and the directrix y = 4.

The graphical representation of the parabola is seen in the image attached below.

Learn more about the graph of a parabolic equation here:

brainly.com/question/12896871

#SPJ1

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Find the area of quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21)

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Step-by-step explanation:

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Quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21).

Now, to find the area.

The coordinates of the quadrilateral are A(1,1), B(7,-3), C(12,2), D(7,21).

So, to find the area we need to bisect the quadrilateral ABCD and get the triangles ABC and ADC and then calculate their areas:

In Δ ABC:

$A(x_1,y_1)=(1,1)\:,\:B(x_2,y_2)=(7,-3)\:and\:C(x_3,y_3)=(12,2)$

Now, to get the area of triangle ABC:

$Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-3-2)+(7)(2-1)+12(1--3)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-5)+(7)(1)+12(4)\right|$

On solving we get:

$Area\,of\,triangle\,=25.$

In Δ ADC:

$A(x_1,y_1)=(1,1)\:,\:D(x_2,y_2)=(7,21)\:and\:C(x_3,y_3)=(12,2)$

Now, to get the area of triangle ADC:

$Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(21-2)+(7)(2-1)+12(1-21)\right|$

On solving it by same process as above we get:

$Area\,of\,triangle\,=114$

Now, to get the area of the quadrilateral we add the areas of the triangles ABC and ADC:

$25+114\\=25+114\\=139\ unit^2$

Therefore, area of quadrilateral ABCD is 139 unit^2.

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3 years ago