# Find a function f whose graph is a parabola with the given vertex and that passes through the given point. vertex (−1, 8); point (−2, −2)?

Find a function f whose graph is a parabola with the given vertex and that passes through the given point.

vertex

(−1, 8); point

(−2, −2)

f(x)=

### 3 Answers

- Wayne DeguManLv 72 months ago
f(x) = ax² + bx + c

Using calculus, we can say f '(x) = 2ax + b

Now, f '(x) = 0 at vertex, i.e. when x = -1

so, -2a + b = 0

or, b = 2a

Using (-1, 8) we have, a(-1)² + b(-1) + c = 8

i.e. a - b + c = 8....(1)

Using (-2, -2) we have, a(-2)² + b(-2) + c = -2

i.e. 4a - 2b + c = -2...(2)

(2) - (1) => 3a - b = -10

Using b = 2a we get:

3a - 2a = -10

so, a = -10 => b = -20 => c = -2

Hence, f(x) = -10x² - 20x - 2 => -2(5x² + 10x + 1)

or, f(x) = -10(x + 1)² + 8

Note: a long method, but just using a different approach.

:)>

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- PinkgreenLv 72 months ago
Let y=ax^2+bx+c be the equation required, then

(-1,8) is on the curve=>

a-b=8-c------(1)

(-2,-2) is on the curve=>

4a-2b=-2-c------(2)

Solving (1) & (2) for a, b get

a=(c-18)/2

b=(3c-34)/2

=>

y=(c-18)(x^2)/2+(3c-34)x/2+c

=>

y '=(c-18)x+(3c-34)/2

y '=0=>x=(34-3c)/[2(c-18)]

x=-1=>(34-3c)/[2(c-18)]=-1=>c=-2

Thus, a=-10; b=-20

=>

y=-10x^2-20x-2

=>

y=-2(5x^2+10x+1) is the required equation.

Check:

y=-2(5x^2+10x+1)=>

y=-10(x^2+2x+1/5)=>

y=-10(x^2+2x+1-1)-2=>

y=-10(x+1)^2+8=>

(-1, 8) is the vertex.

- PinkgreenLv 72 months agoReport
The advantage of this method is that we need not remember the "formula". Just solve the problem

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- L. E. GantLv 72 months ago
First, the vertex equation:

y = a(x+1)^2 + 8

now find a:

-2 = a(-2+1)^2 + 8

==> a(-1)^2 = -10

==> a = -10

so

y = -10(x+1)^2 + 8

Finished..... unless you want to express the equation in general quadratic form

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