Solve SAT quadratic questions with Desmos

Topic: Quadratics
By: SolvePlot

Graph the quadratic, then choose the feature that matches the question: x-intercepts for real roots, an intersection for an equation, or the vertex for an unrestricted minimum or maximum. Check domain restrictions before using the vertex.

Original worked examples

A quadratic graph is a parabola. Its $x$ -intercepts show where its output is zero. Its vertex gives its turning point. If the leading coefficient is positive, the vertex is the unrestricted minimum; if it is negative, the vertex is the unrestricted maximum.

Keep the question's target in view. A question asking for the minimum value wants the $y$ -coordinate, not the $x$ -coordinate where that minimum occurs.

Should you read a root or the vertex?

For $f \left(x\right) = 0\text{,}$ look where the curve meets the $x$ -axis. For the least value of $f \left(x\right)\text{,}$ inspect the lowest permitted point of the curve. These answer different questions even though they use the same graph.

A domain restriction can change the minimum. If the vertex is outside the allowed interval, check the permitted boundary and how the curve behaves on that interval.

Original example

Work through the question

For $f(x)=x^2-8x+12\text{,}$ what is the least possible value of $f(x)$ over all real $x\text{?}$

Answer $-4$

The parabola x² − 8x + 12 has vertex (4, −4) and roots 2 and 6. Illustration; follow the rows below in Desmos.

What to enter in Desmos

Use one expression row for each entry. Underscores create subscripts; the tilde ~ fits parameters.

```
f(x)=x^2-8x+12
```
Graph the given function. A window from x = 0 to 8 and y = -6 to 14 includes the useful features.

Read the result

Click the lowest point of the curve to inspect the vertex (4, -4). The minimum value is the second coordinate, -4. The first coordinate, 4, tells you where it occurs.

Check it

$f(4)=16-32+12=-4\text{.}$ Also, $f(x)=(x-4)^2-4\text{;}$ a square cannot be negative, so no real input produces a smaller value.

Try the entries in Desmos

The algebraic solution

Complete the square: $x^2-8x+12=(x-4)^2-4\text{.}$
Since $(x-4)^2$ is at least $0\text{,}$ the least output is $-4\text{,}$ attained when $x=4\text{.}$

When is each method useful?

Use Desmos when

A graph can make roots, intersections, and turning points easy to inspect without first transforming the expression. It is useful when the required value can be read clearly and checked.

Use algebra when

Factoring or completing the square can be shorter for convenient coefficients. It also proves an exact minimum and handles symbolic questions that a rounded coordinate cannot settle.

Common mistakes to avoid

Do not report 4 when the question asks for the minimum output; here the answer is -4.
The roots are 2 and 6 in this example. Neither is the minimum value.
Check any interval or integer restriction. For x ≥ 5 here, the minimum is f(5) = -3, not -4.
A graph that seems not to touch an axis may need a different window. Use the equation and conditions to confirm your conclusion.

Try one yourself

Find the real solutions of $x^2-8x+12=0$

See the answer and steps

$x=2$ or $x=6$

Graph the equation or inspect the $x$ -intercepts of the original function. Algebra gives $(x-2)(x-6)=0\text{,}$ so the roots are $2$ and $6\text{.}$ Both make the original expression zero.

Desmos entries:

x^2-8x+12=0

Calculator references

The examples above are original. These official resources explain the calculator features and test-day rules.

Desmos: graphing calculator user guide