Five facts, then the curve draws itself — and symmetry hands you every second point.
Find the vertex
y = x² − 6x + 5
The axis of symmetry is x = −b2a = 62 = 3, and substituting back gives y = 9 − 18 + 5 = −4. The vertex is (3, −4). B drops the minus in −b2a — since b is −6, −b comes out as +6.
opens upward ∪
the vertex is the lowest point
crosses the y-axis at 5
one point, for free
c is always the y-intercept. Put x = 0 into the equation and the first two terms vanish — nothing is left but c.
−b2a gives the x. The y needs one substitution.
x = −b2a = −−62
b is negative six
x = 3
the axis of symmetry
y = 3² − 6(3) + 5
put x = 3 back in
y = 9 − 18 + 5 = −4
vertex (3, −4)
The axis of symmetry is a line, so it is written x = 3, not just 3. The vertex is a point, (3, −4). Questions ask for one or the other and the choices include both.
The same two lines run backwards. If y = x² − 2x + k has a minimum value of 5, the axis is x = 1 and y = 1 − 2 + k = k − 1 = 5, so k = 6. If the axis of y = ax² + 10x + 1 is x = −1, then −10 ÷ 2a = −1, so a = 5 and the curve opens upward.
Crossing the x-axis means y = 0 — which is just solving the quadratic.
x² − 6x + 5 = 0
set y to zero
(x − 1)(x − 5) = 0
two numbers: ×5, +(−6)
x = 1 or x = 5
the x-intercepts
| Discriminant | The graph |
|---|---|
| b² − 4ac > 0 | cuts the x-axis twice |
| b² − 4ac = 0 | just touches it — the vertex sits on the axis |
| b² − 4ac < 0 | misses it completely, no real roots |
Here 36 − 20 = 16, positive — two crossings, exactly as we found.
Every point has a twin the same distance on the other side of the axis.
(0, 5) is 3 left of x = 3
the y-intercept
3 right of the axis is x = 6
same distance
so (6, 5) is on the curve
no substitution needed
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y | 0 | −3 | −4 | −3 | 0 |
The row reads the same backwards. That is the symmetry showing up in the table, and it halves the arithmetic.
A parabola has no corners, and its lowest point is at the vertex — not at whichever plotted point happens to be lowest. Sketch through the vertex smoothly and let the two arms rise at the same rate.
The shape is always the same curve — a only stretches it.
| a | Effect |
|---|---|
| a > 1 | narrower — it climbs faster |
| a = 1 | the standard y = x² shape |
| 0 < a < 1 | wider and flatter |
| a < 0 | the whole thing flips upside down |
Our curve is y = (x − 3)² − 4. Double the whole right-hand side and you get y = 2(x − 3)² − 8: same axis, same roots at 1 and 5, but a narrower curve with the vertex twice as deep. Change only the x² term and the axis −b2a shifts as well.
Three things to read: the opening, the y-intercept, and which side the vertex is on.
| Look at | It tells you |
|---|---|
| opens up / opens down | a > 0a < 0 |
| crosses the y-axis above / below 0 | c > 0c < 0 |
| vertex left of the y-axis | a and b have the same sign |
| vertex right of the y-axis | a and b have opposite signs |
| vertex on the y-axis | b = 0 |
opens upward
a > 0
y-intercept below 0
c < 0
vertex x = −b2a is negative
left of the y-axis
−b2a < 0 with a > 0
so −b < 0 and b > 0
Why the side rule works: the vertex sits at x = −b2a, which is negative exactly when a and b share a sign.
The same five facts each time; what changes is how many x-intercepts there are.
a negative: y = −x² + 4x + 5
x = −4 ÷ (2 · −1) = 2
axis x = 2
y = −4 + 8 + 5 = 9
vertex (2, 9), the highest point
x² − 4x − 5 = 0, so (x − 5)(x + 1) = 0
y = 0, multiplied by −1
x-intercepts −1 and 5; y-intercept 5; mirror point (4, 5)
(0, 5) is 2 left of the axis
The steps are unchanged; the curve is an ∩ with the vertex on top, and the roots −1 and 5 are again 3 either side of x = 2.
−4 ÷ (−2) is +2. Dropping one of the minus signs puts the axis at x = −2, on the wrong side of the y-axis.
No x-intercepts: y = x² − 4x + 7
x = 2, y = 4 − 8 + 7 = 3
vertex (2, 3), above the axis
D = 16 − 28 = −12
negative: never meets the x-axis
(0, 7) and its mirror (4, 7); (1, 4) and (3, 4)
x = 1: 1 − 4 + 7 = 4
With no roots to plot, the table supplies the extra points. A vertex above the axis on an upward curve already tells us there are no crossings.
Touching the axis: y = x² + 6x + 9
x = −6 ÷ 2 = −3, y = 9 − 18 + 9 = 0
vertex (−3, 0)
D = 36 − 36 = 0
one x-intercept, at the vertex
(0, 9) and its mirror (−6, 9)
3 either side of x = −3
The curve rests on the x-axis at one point, because y = (x + 3)². The vertex and the root are the same point.
Intercepts give factored form, a vertex gives vertex form — then expand.
y = a(x + 1)(x − 4)
roots −1 and 4
−4 = a(1)(−4), so a = 1
use (0, −4)
y = x² − 3x − 4
expand
y = (x + 2)² − 1
vertex (−2, −1), with a = 1
y = x² + 4x + 4 − 1
expand the square
y = x² + 4x + 3
so b = 4 and c = 3
The bracket shows the opposite sign. Writing (x − 2)² − 1 for a vertex at (−2, −1) flips b to −4.
A vertex plus one more point fixes a. Vertex (2, −9) through (0, −5): −5 = a(0 − 2)² − 9, so 4a = 4 and a = 1, giving y = (x − 2)² − 9. Its x-intercepts: (x − 2)² = 9, x − 2 = ±3, so x = 5 or x = −1.
Graph items show a picture and ask for the matching equation, the signs of a, b and c, or one coefficient. Test the choices against three free facts — the y-intercept c, the opening direction, and the axis midway between the x-intercepts — and most choices fall away in seconds.
The problem
For y = −x² + 4x − 7, which way does it open and how many times does it cross the x-axis?
a = −1 is negative, so it opens downward. The discriminant is 4² − 4(−1)(−7) = 16 − 28 = −12, which is negative, so there are no real roots and it never reaches the axis. Watch −4ac: two negatives inside make it negative overall here.
