Describe the graph of the quadratic from the original y = x^2 .

Mathematics

1 answer:

Citrus2011 [14]2 years ago

4 0

Answer:

a. Narrower

b. Shifts left

c. Opens up

d. Shifts up

Step-by-step explanation:

The original quadratic equation is y = x²

The given quadratic equation is y = 5·(x + 4)² + 7

The given quadratic equation is of the form, f(x) = a·(x - h)² + k

a. A quadratic equation is narrower than the standard form when the coefficient is larger than the coefficient in the original equation

The quadratic coefficient is 5 > 1 in the original, therefore, the quadratic equation is <em>narrower</em>

b. Given that the given quadratic equation has positive 'a', and 'b', and h = -4, therefore, the axis of symmetry <em>shifts left</em>

c. The quadratic coefficient is positive, (a = 5), therefore, the quadratic equation <em>opens down</em>

d. The value of 'k' gives the vertical shift, therefore, the given quadratic equation with k = 7, <em>shifts up.</em>

You might be interested in

A circle is centered at the point (5, -4) and passes through the point (-3, 2).

zepelin [54]

Answer:

(x-5)^2+(y+4)^2=100

Step-by-step explanation:

As we know the given points

Center = (5, -4)

and

Point on circle = (-3,2)

The distance between point on circle and center will give us the radius of circle

So,

The formula for distance is:

$\sqrt{(x_{2}-x_{1} )^{2}+(y_{2}-y_{1})^{2}}\\Taking\ center\ as\ point\ 1\ and\ the\ other\ point\ as\ point\ 2\\d=\sqrt{(-3-5)^{2}+(2-(-4))^{2}}\\d=\sqrt{(-8)^{2}+(2+4)^{2}}\\d=\sqrt{(-8)^{2}+(6)^{2}}\\\\d=\sqrt{64+36}\\d=\sqrt{100} \\ d=10\\So\ the\ radius\ is\ 10$