I rlly need help :( this is hard

Mathematics

1 answer:

GrogVix [38]3 years ago

8 0

Answer:

Step-by-step explanation:

The domain does not exist when the denominater of an equation is zero. So in this case if x was -2 and we added 2 the denominator would be zero, and that value does not exist in the domain. Hope this helps!

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Step-by-step explanation:

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Use the Parabola tool to graph the quadratic function. f(x)=2x^2−12x+19 Graph the parabola by first plotting its vertex and then

Debora [2.8K]

To find the vertex of our parabola, we are going to use the vertex formula: For a quadratic of the form $f(x)=x^2+bx+c$ it vertex $(h,k)$ is given by $h= \frac{-b}{2a}$ , $k=f(h)$ .
We can infet from our parabola that $a=2$ and $b=-12$ . So lets replace those values in our formula:
$h= \frac{-b}{2a}$
$h= \frac{-(-12)}{2(2)}$
$h= \frac{12}{4}$
$h=3$
$k=f(h)=2(3)^2-12(3)+19$
$k=2(9)-36+19$
$k=18-17$
$k=1$
The vertex $(h,k)$ of our parabola is $(3,1)$

Now to find our second point, we are going to evaluate our parabola at $x=0$
$f(0)=2(0)^2-12(0)+19$
$f(0)=0+0+19$
$f(0)=19$
Our second point is $(0,19)$

We can conclude that we can graph the parabola $f(x)=2x^2-12x+19$ using its vertex $(3,1)$ and the point $(0,19)$ as follows: