Question

Find a formula for the quadratic function whose graph isshown.

Short answer

The required quadratic equation is \(f\left( x \right) = - {x^2} - 2.5x + 1\).

Step-by-step solution

Observe the given graph

It is known that the quadratic function is of the form \(f\left( x \right) = a{x^2} + bx + c\).

The given graph has the\(y\)intercept of 1, so\(f\left( 0 \right) = 1\).

Also, the points \(\left( { - 2,2} \right)\) and \(\left( {1, - 2.5} \right)\) exist on the curve of the given graph. So, the points must satisfy the quadratic function.

Obtain the intercept

Use \(f\left( 0 \right) = 1\) to find the value of \(c\) as shown below:

\(\begin{aligned}f\left( 0 \right) &= 1\\a{\left( 0 \right)^2} + b\left( 0 \right) + c &= 1\\c &= 1\end{aligned}\)

Thus, \(c = 1\).

Obtain the equations

Plug-in points \(\left( { - 2,2} \right)\)and \(\left( {1, - 2.5} \right)\) in the function \(f\left( x \right)\) as shown below:

As the points\(\left( { - 2,2} \right)\)and\(\left( {1, - 2.5} \right)\)satisfies\(f\left( x \right) = a{x^2} + bx + c\). So,\(f\left( { - 2} \right) = 2\)and \(f\left( 1 \right) = - 2.5\).Thus,

\(\begin{aligned}f\left( { - 2} \right) &= 2\\a{\left( { - 2} \right)^2} + b\left( { - 2} \right) + 1 &= 2\\4a - 2b &= 2 - 1\\4a - 2b &= 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\end{aligned}\)

And,

\(\begin{aligned}f\left( 1 \right) &= - 2.5\\a{\left( 1 \right)^2} + b\left( 1 \right) + 1 &= - 2.5\\a + b &= - 2.5 - 1\\a + b &= - 3.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\end{aligned}\)

Solve the equations for \(a,\,b\)

Multiply the second equation by 2 and add it to the first equation to find the value of \(a\) as shown:

\(\begin{aligned}4a - 2b + 2\left( {a + b} \right) &= 1 - 2\left( {3.5} \right)\\6a &= - 6\\a &= - 1\end{aligned}\)

Plugin \(a = - 1\) in any of the equations, we can calculate \(b\)as shown:

\(\begin{aligned}4\left( { - 1} \right) - 2b &= 1\\ - 2b &= 5\\b &= - 2.5\end{aligned}\)

So, \(a = - 1\) and \(b = - 2.5\).

Substitute the values of \(a,\,b,c\) into \(f\left( x \right)\)

On plugging the values of\(a,b,c\) into \(f\left( x \right) = a{x^2} + bx + c\),we get the required quadratic equation as \(f\left( x \right) = - {x^2} - 2.5x + 1\).

Thus, the quadratic equation is \(f\left( x \right) = - {x^2} - 2.5x + 1\).