Prove that y=2x^2-6x+7 and y=x^2-2x+3 touch each other and find the point of contact

Mathematics

1 answer:

zepelin [54]2 years ago

3 0

Answer:

See below ~

Step-by-step explanation:

We need to prove they are equal.

Equate the values to each other.

2x² - 6x + 7 = x² - 2x + 3
x² - 4x + 4 = 0
(x - 2)² = 0
x = 2

Now input the x-value in the 1st equation.

y = 2(2)² - 6(2) + 7
y = 8 - 12 + 7
y = -4 + 7
y = 3

Now input the same in the 2nd equation.

y = (2)² - 2(2) + 3
y = 4 - 4 + 3
y = 3

Since the same x-value, on inputting in both equations, give the same y-value, it is proved they touch each other.

Point of Contact

(x, y)
(2, 3)

You might be interested in

WHAT IS ANGLE? Need it right know

guapka [62]

Answer:

No file attached but if its a triangle its 180 Degrees in total

if its a cirle its 360 Degrees in total if its a square and other geometrical figures its 360 Degrees

Step-by-step explanation:

Would love to see what we have to calculate

5 0

3 years ago

Read 2 more answers

Simplify exponential form

wolverine [178]

Answer:

$( \frac{5}{4})^{6}$

Step-by-step explanation:

$\frac{4}{5} \div \bigg[ {\bigg( \frac{4}{5} \bigg)}^{7} .{\bigg( \frac{4}{5} \bigg)}^{0}\bigg] \\ \\ = \frac{4}{5} \div \bigg[ {\bigg( \frac{4}{5} \bigg)}^{7} .1\bigg] \\ \\ = \frac{4}{5} \div {\bigg( \frac{4}{5} \bigg)}^{7}\\ \\ = \frac{4}{5} \times {\bigg( \frac{5}{4}\bigg)}^{7} \\ \\ = {\bigg( \frac{5}{4}\bigg)}^{6}$