You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equation of Tangent at a Point (x,y) in Terms of f'(x)

Find the valu...

Question

Find the value of $a + b + c$ , if points $(- 3, 10), (0, 1)$ and $(2, 15)$ lie on the curve $y = a x^{2} + b x + c$ .

6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $6$
$y = a x^{2} + b x + c$
Three points $(- 3, 10), (0, 1), (2, 15)$ lie on the curve.
Substituting them into the curve, we get
$10 = 9 a - 3 b + c . . . (1)$
$1 = c . . . (2)$
$15 = 4 a + 2 b + c . . . (3)$
Substituting value of c from equation 2 in equations 1 and 3, we have
$10 = 9 a - 3 b + 1$ or $3 = 3 a - b . . . (4)$
$15 = 4 a + 2 b + 1$ or $7 = 2 a + b . . . (5)$
Adding equations (4) and (5), we have
$10 = 5 a$ or $a = 2$
And therefore, $b = 3$
$∴ a + b + c = 2 + 3 + 1 = 6$

Suggest Corrections

Similar questions

(- 3, 10), (0, 1)

(2, 15)

y = a x^{2} + b x + c

a + b + c

y = a x^{2} + b x + c = 0

y

6

(\frac{5}{2}, \frac{49}{4}),

a + b + c

y = a x^{2} + b x + c, x \in R,

(1, 2)

y = x,

a, b, c

y = a x^{2} + b x + c

(0, 3), (1, - 4)

(- 1, 4)

c - (a + b) =

y = a x^{2} + b x + c

y

6

(\frac{5}{2}, \frac{49}{4}),

a + b + c

Join BYJU'S Learning Program