Parabola: vertex and intercepts with the axes

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Given the parabola $y=\dfrac{x^2}{2}+3x+4$:

• (a) find the coordinates of the vertex;
• (b) find the intercepts with the $x$-axis;
• (c) find the intercept with the $y$-axis;
• (d) sketch the parabola.

Solution

The coefficients are $a=\dfrac12$, $b=3$, $c=4$. Since $a>0$ the parabola opens upward.

(a) Vertex. $x_V=-\frac{b}{2a}=-\frac{3}{2\cdot\frac12}=-3$ $y_V=\frac{(-3)^2}{2}+3(-3)+4=\frac92-9+4=-\frac12$ So $V\left(-3;\,-\dfrac12\right)$.

(b) Intercepts with the $x$-axis (set $y=0$): $\frac{x^2}{2}+3x+4=0 \;\Longrightarrow\; x^2+6x+8=0 \;\Longrightarrow\; (x+2)(x+4)=0$ hence $x=-4$ and $x=-2$: the points are $(-4;\,0)$ and $(-2;\,0)$.

(c) Intercept with the $y$-axis (set $x=0$): $y=\frac{0}{2}+0+4=4 \;\Longrightarrow\; (0;\,4)$

(d) Graph. Parabola opening upward, vertex at $V\left(-3;-\dfrac12\right)$, crossing the $x$-axis at $(-4;0)$ and $(-2;0)$ and the $y$-axis at $(0;4)$. $\boxed{V\left(-3;-\tfrac12\right)\quad (-4;0),\ (-2;0)\quad (0;4)}$