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scoray [572]
3 years ago
11

Write the equation of the quadratic function whose graph passes through

ormula1" title="(-3,2)" alt="(-3,2)" align="absmiddle" class="latex-formula">, (-1,0), and (1,6).
Mathematics
1 answer:
blagie [28]3 years ago
5 0

Answer:

f(x)=x^2+3x+2

Step-by-step explanation:

We want to write the equation of a quadratic whose graph passes through (-3, 2), (-1, 0), and (1, 6).

Remember that the standard quadratic function is given by:

f(x)=ax^2+bx+c

Since it passes through the point (-3, 2). This means that when x=-3, f(x)=f(-3)=2. Hence:

f(-3)=2=a(-3)^2+b(-3)+c

Simplify:

2=9a-3b+c

Perform the same computations for the coordinates (-1, 0) and (1, 6). Therefore:

0=a(-1)^2+b(-1)+c \\ \\0=a-b+c

And for (1, 6):

6=a(1)^2+b(1)+c\\\\ 6=a+b+c

So, we have a triple system of equations:

\left\{        \begin{array}{ll}            2=9a-3b+c &\\           0=a-b+c \\6=a+b+c        \end{array}    \right.

We can solve this using elimination.

Notice that the b term in Equation 2 and 3 are opposites. Hence, let's add them together. This yields:

(0+6)=(a+a)+(-b+b)+(c+c)

Compute:

6=2a+2c

Let's divide both sides by 2:

3=a+c

Now, let's eliminate b again but we will use Equation 1 and 2.

Notice that if we multiply Equation 2 by -3, then the b terms will be opposites. So:

-3(0)=-3(a-b+c)

Multiply:

0=-3a+3b-3c

Add this to Equation 1:

(0+2)=(9a-3a)+(-3b+3b)+(c-3c)

Compute:

2=6a-2c

Again, we can divide both sides by 2:

1=3a-c

So, we know have two equations with only two variables:

3=a+c\text{ and } 1=3a-c

We can solve for a using elimination since the c term are opposites of each other. Add the two equations together:

(3+1)=(a+3a)+(c-c)

Compute:

4=4a

Solve for a:

a=1

So, the value of a is 1.

Using either of the two equations, we can now find c. Let's use the first one. Hence:

3=a+c

Substitute 1 for a and solve for c:

\begin{aligned} c+(1)&=3 \\c&=2 \end{aligned}

So, the value of c is 2.

Finally, using any of the three original equations, solve for b:

We can use Equation 3. Hence:

6=a+b+c

Substitute in known values and solve for b:

6=(1)+b+(2)\\\\6=3+b\\\\b=3

Therefore, a=1, b=3, and c=2.

Hence, our quadratic function is:

f(x)=x^2+3x+2

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