Question 145·200 Super-Hard SAT Math Questions·Advanced Math

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The graph of a quadratic function $f$ has $x$ -intercepts at $(1,0)$ and $(5,0)$ . The graph of $y=f(x)$ and the line $y=4x-20$ have exactly one point in common. What is the value of $f(2)$ ?

Your answer

Two x-intercepts give a quadratic's factored form, but leave its multiplier unknown. Here the line passes through one of those intercepts, so one intersection is guaranteed. Use exactly one point in common to make the other possible intersection land at that same point, then evaluate the quadratic in Desmos. Don't assume the line has the same output as the quadratic at every input.

Hints

Hint 1
An x-intercept is a point where the output is zero. Each intercept gives one factor of the quadratic, but an unknown multiplier remains. How can you write $f(x)$ using both factors?
Hint 2
Factor the line's expression, $4x-20$ . It has an $x-5$ factor, so the line and quadratic already meet at the intercept where $x=5$ .
Hint 3
Set the two outputs equal and factor the result. One factor gives the known intersection at $x=5$ . For there to be no second intersection, what must the other factor do at $x=5$ ?

Step-by-step

Make both intersections coincide
Step 1
Write the quadratic from its intercepts
An x-intercept has output $0$ . The intercepts at $x=1$ and $x=5$ give one factor each, so $f(x)=a(x-1)(x-5),\quad a\ne0$ . The multiplier $a$ is unknown; it cannot be zero because $f$ is quadratic.
Step 2
Find the intersection already guaranteed
Factor $4$ from the line's output: $4x-20=4(x-5)$ . At $x=5$ , both this expression and $f(x)$ equal $0$ . So the graphs already share $(5,0)$ .
Step 3
Set the outputs equal
At an intersection, the two graphs have the same output for the same input. So any shared point must satisfy $a(x-1)(x-5)=4(x-5)$ . We need this equation to give only one $x$ -value.
Step 4
Move everything to one side
Subtract $4(x-5)$ from both sides: $a(x-1)(x-5)-4(x-5)=0$ . Don't divide by $x-5$ : doing that would erase the known solution $x=5$ .
Step 5
Expose the possible second intersection
Factor out $x-5$ : $(x-5)[a(x-1)-4]=0$ . A product is zero when either factor is zero. The first gives $x=5$ ; the bracket could give another intersection.
Step 6
Make the second factor give the same point
For exactly one shared point, the bracket must also be zero at $x=5$ . Otherwise, because $a\ne0$ , it gives a different input. Substitute $5$ into the bracket: $a(5-1)-4=0$ . Simplify: $4a-4=0$ . Add $4$ to both sides: $4a=4$ . Type $4a\sim4$ in Desmos; the tilde tells it to find $a$ . Under PARAMETERS, it shows $a=1$ .
Step 7
Evaluate the quadratic at 2
Type $f(x)=a(x-1)(x-5)$ below the regression, then type $f(2)$ . Desmos uses $a=1$ and prints $f(2)=-3$ . That is the quadratic's output at $x=2$ , so Grid in -3.

Question 145·200 Super-Hard SAT Math Questions·Advanced Math

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