Porisma

Worked solution

Find exact intercepts and verify the calculator window

Find the coordinates of the axial intercepts of the curve y=x2+2x−15y=x^2+2x-15.

CJC · Lecture · Example 1

Find the exact intercepts algebraically

Set y=0y=0: x2+2x−15=(x+5)(x−3)=0x^2+2x-15=(x+5)(x-3)=0, giving x=−5x=-5 or 33. Set x=0x=0: y=02+2(0)−15=−15y=0^2+2(0)-15=-15. The axial intercepts are (−5,0)(-5,0), (3,0)(3,0) and (0,−15)(0,-15).

How did this part go?

Find each zero using the calculator

The source gives six steps.

  1. Press Y= and enter x2+2x−15x^2+2x-15 into Y1.

  2. Press 2nd → TRACE, then CALCULATE → 2:zero.

  3. Set a left bound to the left of the first root, using the arrows and ENTER or entering a number.

  4. Set a right bound to its right. The bounds should contain only one root, or the calculator may return the other.

  5. For Guess?, move near that root and press ENTER.

  6. Repeat steps 2–5 for the other root.

The displayed zeros are −5-5 and 33. A numerical result such as −4.9999999-4.9999999 must be checked before rounding: substitution of −5-5 gives y=0y=0.

How did this part go?

Change the window before finding the y-intercept

The source gives five further steps.

  1. Press WINDOW; the illustrated defaults are Xmin=−10-10, Xmax=1010, Ymin=−10-10, Ymax=1010.

  2. Set Ymin=−20-20. Window selection is a trial-and-error process; there is no single required window.

  3. Press GRAPH to display the previously clipped minimum.

  4. Press 2nd → TRACE, then CALCULATE → 1:value.

  5. Enter 0 → ENTER to obtain y=−15y=-15.

Independently, completing the square gives y=(x+1)2−16y=(x+1)^2-16, so the minimum (−1,−16)(-1,-16) explains why the original Ymin hides it. The reconstruction compares these two windows.

How did this part go?

Choose a method suited to the requested precision

Both copies explicitly require algebraic/analytical working when exact coordinates are requested. When using the calculator, use its zero/value routines; moving along the curve with arrow keys and simply reading coordinates is not accurate enough. The graph supports the calculation, while algebra proves the exact values.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

The same parabola y=x²+2x−15 in two windows. With x and y from −10 to 10 the vertex and y-intercept are clipped; with Ymin −20 both vertex (−1,−16) and y-intercept (0,−15) appear. Both windows show x-intercepts −5 and 3. Dashed line on the expanded panel marks the old Ymin −10.
The same parabola y=x²+2x−15 in two windows. With x and y from −10 to 10 the vertex and y-intercept are clipped; with Ymin −20 both vertex (−1,−16) and y-intercept (0,−15) appear. Both windows show x-intercepts −5 and 3. Dashed line on the expanded panel marks the old Ymin −10.
The same parabola y=x²+2x−15 in two windows. With x and y from −10 to 10 the vertex and y-intercept are clipped; with Ymin −20 both vertex (−1,−16) and y-intercept (0,−15) appear. Both windows show x-intercepts −5 and 3. Dashed line on the expanded panel marks the old Ymin −10.

Common mistakes

A clipped display does not remove the minimum or the y-intercept; verify approximate zeros before rounding them.

Solution notes

Both Student and Teacher originals compared. Printed answers retained; independent explanations identified in the working. No printed marks or external attribution.

Read the hint