Write a standard form of a quadratic equation that has roots - 5/2 and 3.

Mathematics

1 answer:

lora16 [44]3 years ago

6 0

Answer:

2x^2 - x - 15 = 0.

Step-by-step explanation:

In factor form this is

(x + 5/2)(x - 3) = 0

x^2 + 5/2x - 3x - 15/2 = 0

x^2 - 1/2x - 15/2 = 0

Multiply through by 2:

2x^2 - x - 15 = 0.

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$\bf 2[x^2+y^2]^2=25(x^2-y^2)\qquad \qquad 
\begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 3}}\quad ,&{{ 1}})\quad 
\end{array}\\\\
-----------------------------\\\\
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\\\\\\
$

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\\\\\\
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notice... a derivative is just the function for the slope

now, you're given the point 3,1, namely x = 3 and y = 1

to find the "m" or slope, use that derivative, namely $f'(3,1)=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}$

that'd give you a value for the slope

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in the point-slope form

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad
\begin{cases}
x_1=3\\
y_1=1\\
m=slope
\end{cases}\\ \qquad \uparrow\\
\textit{point-slope form}$

and then you solve it for "y", I gather you don't have to, but that'd be the equation of the tangent line at 3,1

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Step-by-step explanation: