E3.4 - Finding a quadratic function (Extended)

E3.4 - Finding a quadratic function (Extended)

Many parabolas have the same lowest point. What extra information picks out just one of them? You will choose a useful quadratic form from a vertex or roots, use another point to find its coefficient, and recognise when a leading coefficient of 1 already fixes the rule.

Vertex and point

A quadratic's vertex is its turning point. The expression (x−2)2(x-2)^2 is smallest when x=2x=2, because a square cannot be negative. Thus y=(x−2)2−3y=(x-2)^2-3 has minimum point (2,−3)(2,-3). Multiplying the square by a positive number changes the steepness but leaves that vertex fixed; a negative multiplier makes it a maximum instead.

Vertex form of a quadratic

y=a(x−h)2+k,a≠0y=a(x-h)^2+k,\qquad a\ne0

The vertex is (h,k)(h,k) and the axis of symmetry is x=hx=h. It is a minimum when a>0a>0, and a maximum when a<0a<0. This form is not in the supplied formula list: learn it.

Use the vertex to write down hh and kk. A second point with a different input from the vertex then fixes aa.

[DIAGRAM: asset_name: A quadratic from its vertex and another point; asset_slug: quadratic_vertex_point; caption: The vertex fixes the brackets and final constant; the second point fixes the multiplier.; description: Exact plot of y equals 2 times x minus 2 squared minus 3, showing vertex 2 comma minus 3, another point 4 comma 5, and dashed symmetry axis x equals 2. The symmetric point zero comma 5 is also marked.]
The vertex fixes the brackets and final constant

Worked example 1: an upward-opening parabola

Question. A quadratic has vertex (2,−3)(2,-3) and passes through (4,5)(4,5). Find its equation.

Plan. Put the vertex into vertex form, then substitute the other point to find aa.

y=a(x−2)2−3use the vertex5=a(4−2)2−3use the other point8=4aadd 3 and square 2a=2\begin{aligned} y&=a(x-2)^2-3 &&\text{use the vertex}\\ 5&=a(4-2)^2-3 &&\text{use the other point}\\ 8&=4a &&\text{add 3 and square 2}\\ a&=2 \end{aligned}

Answer. y=2(x−2)2−3y=2(x-2)^2-3; expanded form y=2x2−8x+5y=2x^2-8x+5 is equivalent.

Check. The square is zero at x=2x=2, giving −3-3, and its positive coefficient confirms a minimum. At the mirror point x=0x=0, the equation gives 5, matching the point at x=4x=4 by symmetry.

Keep the vertex and the second point's horizontal coordinate fixed, but move that point below the vertex. The coefficient must now be negative.

Worked example 2: a downward-opening parabola

Question. A quadratic has vertex (2,−3)(2,-3) and passes through (4,−11)(4,-11). Find its equation.

Plan. Use the same vertex form and substitute the changed output.

−11=a(4−2)2−3−8=4aadd 3a=−2divide by 4\begin{aligned} -11&=a(4-2)^2-3\\ -8&=4a &&\text{add 3}\\ a&=-2 &&\text{divide by 4} \end{aligned}

Answer. y=−2(x−2)2−3y=-2(x-2)^2-3.

Check. Since −2(x−2)2≤0-2(x-2)^2\le0, every output is at most −3-3: the stated vertex is a maximum. At x=4x=4, the output is −2(4)−3=−11-2(4)-3=-11, as required.

Why must the extra point differ from the vertex? Substituting the vertex gives k=a(0)2+kk=a(0)^2+k, true for every aa. It supplies no information about the multiplier. The opening question needs a point away from the turning point.

Watch out

A vertex at (−2,7)(-2,7) gives y=a(x+2)2+7y=a(x+2)^2+7. The bracket is zero at the vertex: check that input instead of copying its sign.

Use a new vertex and another point.

Roots and point

An xx-intercept is a point where the output is zero. If those intercepts are (−1,0)(-1,0) and (3,0)(3,0), the factors x+1x+1 and x−3x-3 build both zeros into the expression. A non-zero multiplier preserves the zeros but changes the other outputs.

Root form of a quadratic

y=a(x−r)(x−s),a≠0y=a(x-r)(x-s),\qquad a\ne0

The roots are rr and ss, giving intercepts (r,0)(r,0) and (s,0)(s,0). If r=sr=s, the repeated factor gives one touching point. This form is not in the supplied formula list: learn it.

A further point that is not an xx-intercept determines aa. Multiplying out is optional unless the question asks for expanded form.

Continue this lesson.

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