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3 years ago

What is the equation of the line that passes through the point (4, 3) and is perpendicular to the line x + y = 4?

1 answer:

5 0

ANSWER

$y - x = - 1$

EXPLANATION

The required equation passes through:

(4,3) and is perpendicular to x+y=4.

Rewrite the given equation in slope-intercept form:

$y = - x + 4$

The slope of this line is -1.

The slope of the required line is perpendicular to this line, so we find the negative reciprocal of this slope.

$m= - \frac{1}{ - 1} = 1$

The equation of the line can be found using:

$y - y_1 = m(x - x_1)$

We substitute the slope and point to obtain:

$y - 3 = 1(x - 4)$

We simplify to get:

$y - 3 = x - 4$

$y - x = - 4 + 3$

The required equation is

$y - x = - 1$

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andreev551 [17]

Answer:

$\textsf{A)} \quad x=-2, \:\:x=\dfrac{5}{2}$

$\textsf{B)} \quad \left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

C) See attachment.

Step-by-step explanation:

Given function:

$f(x)=2x^2-x-10$

To factor a <u>quadratic</u> in the form $ax^2+bx+c$ <em> , </em>find two numbers that multiply to $ac$ and sum to $b$ :

$\implies ac=2 \cdot -10=-20$

$\implies b=-1$

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Rewrite $b$ as the sum of these two numbers:

$\implies f(x)=2x^2-5x+4x-10$

Factor the first two terms and the last two terms separately:

$\implies f(x)=x(2x-5)+2(2x-5)$

Factor out the common term (2x - 5):

$\implies f(x)=(x+2)(2x-5)$

The x-intercepts are when the curve crosses the x-axis, so when y = 0:

$\implies (x+2)(2x-5)=0$

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$\implies (x+2)=0 \implies x=-2$

$\implies (2x-5)=0 \implies x=\dfrac{5}{2}$

So the x-intercepts are:

$x=-2, \:\:x=\dfrac{5}{2}$

The x-value of the vertex is:

$\implies x=\dfrac{-b}{2a}$

Therefore, the x-value of the vertex of the given function is:

$\implies x=\dfrac{-(-1)}{2(2)}=\dfrac{1}{4}$

To find the y-value of the vertex, substitute the found value of x into the function:

$\implies f\left(\dfrac{1}{4}\right)=2\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)-10=-\dfrac{81}{8}$

Therefore, the vertex of the function is:

$\left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

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Plot the vertex found in Part B.

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