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Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s

Nataliya [291]

Answer:

$f(x)=\frac{1}{2}x^2-4x+5$

Step-by-step explanation:

A parabola is written in the form

$f(x)=a((x-h)^2+k)$ (1)

where:

$h$ is the x-coordinate of the vertex of the parabola

$ak$ is the y-coordinate of the vertex of the parabola

$a$ is a scale factor

For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:

$h=4$ (2)

$ak=-3$

From this last equation, we get that $a=\frac{-3}{k}$ (3)

Substituting (2) and (3) into (1) we get the new expression:

$f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3$ (4)

We also know that the parabola contains the point (2,-1), so we can substitute

x = 2

f(x) = -1

Into eq.(4) and find the value of k:

$-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6$

So we also get:

$a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}$

So the equation of the parabola is:

$f(x)=\frac{1}{2}((x-4)^2 -6)$ (5)

Now we want to rewrite it in the standard form, i.e. in the form

$f(x)=ax^2+bx+c$

To do that, we simply rewrite (5) expliciting the various terms, we find:

$f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5$

6 0

3 years ago

the height of a triangle is 4 meters less than the base. if the area of the triangle is 6 m^2 what is the height of the triangle

inna [77]

Answer:

$2m$

Step-by-step explanation:

Alrighty let's do this.

We know that formula for the Area of a triangle is:

$A = \frac{1}{2}bh$

They give us the area as $6m^2$ , so let's include it in the equation.

$6 = \frac{1}{2}bh$

Now we reach a stage where we have 2 unknown variables! That means we can't solve it in its current state. So the idea you should have in cases like these where you have 2 or more unknown variables is, "Can I represent this one variable in terms of another variable?" In this case you can do exactly that. You can represent height in terms of length of the base. We are told the height of the triangle is 4 meters less than the base. That is telling us that $h = b-4$

So replace $h$ in the equation with $b-4$ .

You will now get:

$6 = \frac{1}{2}b(b-4)$

Now we can work towards solving. Let's get simplifying.

$12 = b(b-4)\\12 = b^2 - 4b\\$

bring everything to one side so we can make a quadratic and factor:

$0 = b^2 - 4b - 12\\0 = (b-6)(b+2)\\\\b = 6\\b\neq -2$

We get that $b=6$ .

Since we need the height of the triangle we'll need to call back on what h is. We found earlier that $h = b - 4$ , so to find h, we just sub in our b value into that.