Question 394·Hard·Math

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The graph shows line $L$ and the parabola $y=x^2+bx+c$ , where $b$ and $c$ are constants. The dashed vertical line shown is the axis of symmetry of the parabola, and line $L$ touches the parabola at exactly one point.

Which choice is the value of $c$ ?

A line that touches a parabola once gives a quadratic with one solution when you set their $y$ -values equal. Its discriminant, the expression under the square root in the quadratic formula, must be zero. Read the symmetry axis and two labeled line points to build the equations, then use a Desmos regression to find the missing constant. A near-touch on a graph doesn't establish an exact value.

Hints

Hint 1
The axis of symmetry runs through the parabola's lowest point. For $y=x^2+bx+c$ , its $x$ -coordinate is $-b/2$ . Use the dashed line to find $b$ before working with line $L$ .
Hint 2
A line's slope is its change in $y$ divided by its change in $x$ . Use both labeled points to find the slope, then use the point at $x=0$ to write the line's equation.
Hint 3
At a tangent point, the line and parabola share exactly one $x$ -value. Set their $y$ -expressions equal. What must the resulting quadratic's discriminant be?

Step-by-step

Approach 1: Use the discriminant for one touching point
Step 1
Find the parabola's $x$ -coefficient
The dashed axis of symmetry is $x=2$ . For $y=x^2+bx+c$ , the axis is $x=-b/2$ , so set: $-\frac{b}{2}=2$ . Multiply by $-2$ :
$b=-4.$
Step 2
Find line $L$ 's slope
The labeled points are $(0,1)$ and $(6,4)$ . The slope is the change in $y$ divided by the change in $x$ . Type $\frac{4-1}{6-0}$ in Desmos; it displays $0.5$ , so the slope is $\frac12$ .
Step 3
Write line $L$ 's equation
The point $(0,1)$ is the line's $y$ -intercept: its height where $x=0$ . With slope $\frac12$ , type $y=\frac{x}{2}+1$ ; Desmos graphs the line through the two labeled points. Don't use the line's intercept as $c$ . The $c$ in the question belongs to the parabola.
Step 4
Set the two heights equal
At an intersection, both graphs have the same $y$ -value for the same $x$ . Using $b=-4$ and the line's equation, set their heights equal:
$x^2-4x+c=\frac{x}{2}+1.$
Step 5
Put the quadratic equal to zero
Subtract $\frac{x}{2}+1$ from both sides: $x^2-4x+c-\frac{x}{2}-1=0$ . Combine the $x$ -terms: $x^2-\frac92x+c-1=0$ . This equation's solutions are the shared $x$ -values of the two graphs.
Step 6
Translate touching once into an equation
For $Ax^2+Bx+C=0$ , the discriminant is $B^2-4AC$ . When it is $0$ , the two solutions from the quadratic formula become one. Here $A=1$ , $B=-\frac92$ , and $C=c-1$ . One touching point means the shared-point quadratic has discriminant $0$ . Set:
$\left(-\frac92\right)^2-4(1)(c-1)=0.$
Step 7
Find the exact value of $c$
Type $\left(-\frac92\right)^2-4(c-1)\sim0$ . The $\sim$ asks Desmos to fit $c$ , rather than make a slider; under PARAMETERS it shows $c=6.0625$ . Type $c$ on the next line and use its fraction button to see $\frac{97}{16}$ . Finally, graph $y=x^2-4x+c$ alongside the line: they touch once. So the parabola's $y$ -intercept, $c$ , is $\frac{97}{16}$ . Choice B.

Approach 2: See tangency as a smallest vertical gap
Step 1
Rewrite the gap as a square
Subtract the line's height from the parabola's height using the equations above: $x^2-4x+c-\left(\frac{x}{2}+1\right)=x^2-\frac92x+c-1$ . To complete the square, half of $\frac92$ is $\frac94$ , whose square is $\frac{81}{16}$ . Add and subtract that amount: $x^2-\frac92x+c-1=\left(x-\frac94\right)^2-\frac{81}{16}+c-1$ . Combine the constants: $\left(x-\frac94\right)^2+c-\frac{97}{16}$ . This expression measures the vertical gap between the graphs.
Step 2
Make the smallest gap zero
The square is smallest at $x=\frac94$ , where it equals $0$ . For one touching point, the smallest vertical gap must also be $0$ , so set: $c-\frac{97}{16}=0$ . Add $\frac{97}{16}$ : $c=\frac{97}{16}$ . So the parabola's $y$ -intercept is $\frac{97}{16}$ . Choice B.

Question 394·Hard·Math

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