A line parrell to y=5x+9 and intersects the point (-2,-3). What is the equation of this parrell line

Mathematics

1 answer:

zmey [24]3 years ago

6 0

Answer:

$y - 5x = 4$

Step-by-step explanation:

$y = 5x + 9 \\ m1 = 5 \\ since \: the \: line \: is \: parallel \: m1 = m2 \\ y - y1 = m(x - x1) \\ y - ( - 3) = 5(x - ( - 2)) \\ y + 4 = 5(x + 2) \\ y + 4 = 5x + 10 \\ y - 5x = 10 - 6 \\ y - 5x = 4$

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In f(x) = 2x2 − 8x − 10, the y-intercept is ? at ? and the x-intercepts are (-1, 0) and ? .

IgorLugansk [536]

<h2>1. y-intercept</h2>

$\boxed{(0,-10)}$

The quadratic function $f(x)=2x^22-8x-10$ represents a parabola. In fact, the graph of a quadratic function is a special type of U-shaped curve called a parabola. To find the y intercept, we set $x=0$ as follows:

$f(x)=2x^2-8x-10 \\ \\ If \ x=0 \rightarrow f(0)=-10 \\ \\ Then \ y-intercept: \\ \\ (0,-10)$

<h2>2. x-intercepts</h2>

$\boxed{(-1,0) \ and \ (5,0)}$

To find the other x-intercept, we must set $y=0$ as follows:

$f(x)=2x^2-8x-10 \\ \\ If \ y=0 \rightarrow 2x^2-8x-10=0 \\ \\ Using \ the \ quadratic \ formula: \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \therefore x=\frac{-(-8) \pm \sqrt{(-b)^2-4(2)(-10)}}{2(2)} \\ \\ \therefore x_{1}=-1 \ and \ x_{2}=5$